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```Siyavula textbooks: Grade 11 Maths

Collection Editor:

Free High School Science Texts Project

Collection Editor:

Free High School Science Texts Project

Authors:

Free High School Science Texts Project

Sarah Blyth

Heather Williams

Online:

< http://siyavula.cnx.Org/content/colll243/l.3/ >

CONNEXIONS
Rice University, Houston, Texas

This selection and arrangement of content as a collection is copyrighted by Free High School Science Texts Project.

Collection structure revised: August 3, 2011

PDF generated: August 3, 2011

For copyright and attribution information for the modules contained in this collection, see p. 189.

Exponents 1

Surds 5

Error Margins 11

1 Finance

1.1 Simple depreciation 17

1.2 Compound depreciation 19

1.3 Present and future values 21

1 .4 Finding i and n 22

1.5 Nominal and effective interest rates 24

Solutions 29

2.1 Factorisation 33

2.2 Completing the square 34

2.4 Finding the equation 39

Solutions 43

4 Solving simultaneous equations

4.1 Graphical solution 53

4.2 Algebraic solution 54

Solutions 57

5 Mathematical Models 59

6 Quadratic Functions and Graphs 67

7 Hyperbolic Functions and Graphs 75

8 Exponential Functions and Graphs 81

9 Gradient at a Point 87

10 Linear Programming 93

11 Geometry

11.1 Polygons 103

11.2 Triangle geometry 106

11.3 Co-ordinate geometry 112

11.4 Transformations 116

Solutions 122

12 Trigonometry

12.1 Graphs of trig functions 125

12.2 Trig identities 133

12.3 Reduction formulae 137

12.4 Equations 142

12.5 Cosine and sine identities 148

Solutions 155

13 Statistics

13.1 Standard deviation and variance 161

13.2 Graphical representation of data 166

13.3 Distribution of data 168

13.4 Misuse of statistics 173

IV

Solutions 177

14 Independent and Dependent Events 179

Glossary 185

Index 187

Exponents 1

Introduction

In Grade 10 we studied exponential numbers and learnt that there were six laws that made working with
exponential numbers easier. There is one law that we did not study in Grade 10. This will be described
here.

Laws of Exponents

In Grade 10, we worked only with indices that were integers. What happens when the index is not an integer,
but is a rational number? This leads us to the final law of exponents,

o^ = \/o™ (1)

Exponential Law 7: a« = \/a™

We say that x is an nth root of b if x n = b and we write x = \fb. n th roots written with the radical symbol,
./~, are referred to as surds. For example, (—1) = 1, so —1 is a 4th root of 1. Using law 6, we notice that

( a f)™ = a" x ™ = a m (2)

therefore o^ must be an nth root of a m . We can therefore say

a^ = vV" (3)

For example,

25 = \/¥ (4)

A number may not always have a real nth root. For example, if n = 2 and a = — 1, then there is no real
number such that x 2 = — 1 because x 2 > for all real numbers x.

aside: There are numbers which can solve problems like x 2 = —1, but they are beyond the scope
of this book. They are called complex numbers.

It is also possible for more than one nth root of a number to exist. For example, (—2) = 4 and 2 2 = 4, so
both —2 and 2 are 2nd (square) roots of 4. Usually, if there is more than one root, we choose the positive
real solution and move on.

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Exercise 1: Rational Exponents

Simplify without using a calculator:

4-1 _ 9-

(5)

Exercise 2: More rational Exponents

Simplify:

(16a; 4 ) * (6)

The following videos work through two examples of simplifying expressions.

Khan academy video on exponents - 1

Figure 1

Khan academy video on exponents - 2

Figure 2

Applying laws

Use all the laws to:

1. Simplify:

a. (x°) + 5a; -(0,25r ' 5 + 8l

U i . I

D. S2 — S3

„ 12m 9
C TT

8m 9

d. (64m 6 ) 2%

2. Re-write the following expression as a power of x:

x\ x\ xy x\fx (7)

Exponentials in the Real- World

In Grade 10 Finance, you used exponentials to calculate different types of interest, for example on a savings
account or on a loan and compound growth.

Exercise 3: Exponentials in the Real world

A type of bacteria has a very high exponential growth rate at 80% every hour. If there are 10
bacteria, determine how many there will be in 5 hours, in 1 day and in 1 week?

Exercise 4: More Exponentials in the Real world

A species of extremely rare, deep water fish has an extremely long lifespan and rarely have children.
If there are a total 821 of this type of fish and their growth rate is 2% each month, how many will
there be in half of a year? What will the population be in 10 years and in 100 years?

End of chapter Exercises

1. Simplify as far as possible:

a.

8" 3

b.

\/l6 + 8-f

2.

Simplify:

a.

(x 3 )*

b.

(* 2 )\

c.

(m 5 ) »

d.

{-m 2 Y

c.

-(m 2 ) 1

f.

( 3y t) 4

3.

Simplify as much as

you can:

4. Simplify as much as you can:

5. Simplify as much as you can:

6. Simplify:

7. Simplify:

3a~ 2 b 15 c- 5

(8)

(a- 4 b 3 c)^
(9a 6 b 4 ) * (9)

a*b*\ (10)

X 3 y/x~ (11)

Vx^ (12)

8. Re-write the following expression as a power of x:

x\ xy X\J X\fx
V V (13)

Surds

Surd Calculations

There are several laws that make working with surds (or roots) easier. We will list them all and then explain
where each rule comes from in detail.

y/avb = yob

(1)

V b ~ y%
ifa™ = a ^

Surd Law 1: tfay/b = \fab

It is often useful to look at a surd in exponential notation as it allows us to use the exponential laws we
learnt in Grade 10. In exponential notation, vfa = a™ and \fb = 6« . Then,

i , i

yfa \fb = o *» b <■

= (o6)» (2)

= \/ab

Some examples using this law:

1. ^I6x \/4= ^64 = 4

2. a/2 x V3~2 = \/M = 8

3. VaW x VWd^ = V^b^? = ab 4 c 2

Surd Law 2: ^/f

b Vb

If we look at ^/| in exponential notation and apply the exponential laws then,

\/a

Vb
Some examples using this law:

1. Vl2-=-\/3= \fl= 2

2. s /24 4- y/3= ^8 = 2

3. V^Ts ^ VF = VoW = ab 4

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(3)

Surd Law 3: \/a™ = a*

If we look at \/a m in exponential notation and apply the exponential laws then,

For example,

( °'T «)

V¥ = 2*

= 2i (5)

= V2

Like and Unlike Surds

Two surds >/a and v6 are called iiice surds if m = n, otherwise they are called unlike surds. For example
\/2 and \/3 are like surds, however \/2 and \/2 are unlike surds. An important thing to realise about the
surd laws we have just learnt is that the surds in the laws are all like surds.

If we wish to use the surd laws on unlike surds, then we must first convert them into like surds. In order
to do this we use the formula

^m = Vo&™ (6)

to rewrite the unlike surds so that bn is the same for all the surds.

Exercise 5: Like and Unlike Surds

Simplify to like surds as far as possible, showing all steps: \/3 x \/5

Simplest Surd form

In most cases, when working with surds, answers are given in simplest surd form. For example,

50 = V25 x 2

25 x V2 (7)

h^/2

5\/2 is the simplest surd form of \/50-

Exercise 6: Simplest surd form

Rewrite \/l8 in the simplest surd form:

Exercise 7: Simplest surd form

Simplify: ^/I47+^/I08

This video gives some examples of simplifying surds.

Khan academy video on surds - 1

Figure 1

Rationalising Denominators

It is useful to work with fractions, which have rational denominators instead of surd denominators. It is
possible to rewrite any fraction, which has a surd in the denominator as a fraction which has a rational
denominator. We will now see how this can be achieved.

Any expression of the form y/a + Vb (where a and b are rational) can be changed into a rational number
by multiplying by yfa — Vb (similarly \fa — vb can be rationalised by multiplying by yfa + Vb) . This is
because

(v^+v^) (y/a-Vb) =a-b (8)

which is rational (since a and b are rational).

If we have a fraction which has a denominator which looks like V a + Vb, then we can simply multiply
both top and bottom by y/a — Vb achieving a rational denominator.

c _ y/a—y/b c

s/a+y/b \/a—Vb ^Ja+Vb /g-i

Cy/a—cy/b

a—b

or similarly

c y/a+Vb c

x/a—Vb y/a+Vb y/a—Vb (10)

_ cy/a+cVb ^ '

Exercise 8: Rationalising the Denominator

Rationalise the denominator of: 5x ~} 6

Exercise 9: Rationalising the Denominator

Rationalise the following: 5 ^-~\q

Exercise 10: Rationalise the denominator

Simplify the following: y p +5

The following video explains some of the concepts of rationalising the denominator.

Khan academy video on surds - 2

Figure 2

End of Chapter Exercises

1. Expand:

2. Rationalise the denominator:

3. Write as a single fraction:

4. Write in simplest surd form:

a. a/72 _

b. V45 + V80

r v48

d.

e 4

e - (7IW2)

f 16

(V2CK-v / 12)

5. Expand and simplify:

y/x-V2) (y/E+V2) (11)

(12)

10

^-i

11 fi (13)

2^

(2 + V2)

6.

Expand and simplify:

(2 + V2) (l + y/8\

7.

Expand and simplify:

(l + v^j) (1 + V8 + V3

8.

Rationalise the denominator:

j/-4

yy-2

9.

Rationalise the denominator:

2x-20

yy-Vio

10. Prove (without the use of a calculator) that:

V3 V3 V6 2V3

11. Simplify, without use of a calculator:

98-V8

(14)

(15)

(16)

(17)

(18)

S -5Jj-Ji = ?J? (10)

o()

(20)

Vbl V / 45 + 2\/80

12. Simplify, without use of a calculator:

13. Write the following with a rational denominator:

V5 + 2

14. Simplify:

V98x 6 + V128.T 6

(21)

(22)

(23)

15. Evaluate without using a calculator: 12—^) • ( 2 + 2 )

16. The use of a calculator is not permissible in this question. Simplify completely by showing all your

steps: 3~

12+ ^/(3V3)

17. Fill in the blank surd-form number which will make the following equation a true statement: — 3\/6 x
-2^24 = -a/18 x

10

Error Margins

We have seen that numbers are either rational or irrational and we have see how to round-off numbers.
However, in a calculation that has many steps, it is best to leave the rounding off right until the end.

For example, if you were asked to write 3-\/3 + \/l2 as a decimal number correct to two decimal places,
there are two ways of doing this.

Method 1

3V3 + \/l2

Method 2

3\/3

12

3V3+\/4~3

3^3+2^3

5\/3

5 x 1,732050808.

8,660254038...

8,66

3 x 1,73 + 3,46

5,19+3,46

8,65

(1)

(2)

In the example we see that Method 1 gives 8,66 as an answer while Method 2 gives 8,65 as an answer. The
answer of Method 1 is more accurate because the expression was simplified as much as possible before the

In general, it is best to simplify any expression as much as possible, before using your calculator to work
out the answer in decimal notation.

tip: It is best to simplify all expressions as much as possible before rounding-off answers. This

Exercise 11: Simplification and Accuracy

Calculate \/54 + \/l6. Write the answer to three decimal places.

Exercise 12: Simplification and Accuracy 2

Calculate \/x + 1 + | \/(2x + 2) — (x + 1) if x = 3, 6. Write the answer to two decimal places.

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12

Significant Figures - Extension

In a number, each non-zero digit is a significant figure. Zeroes are only counted if they are between two
non-zero digits or are at the end of the decimal part. For example, the number 2000 has 1 significant figure
(the 2), but 2000, has 5 significant figures. Estimating a number works by removing significant figures from
your number (starting from the right) until you have the desired number of significant figures, rounding as
you go. For example 6, 827 has 4 significant figures, but if you wish to write it to 3 significant figures it
would mean removing the 7 and rounding up, so it would be 6, 83. It is important to know when to estimate
a number and when not to. It is usually good practise to only estimate numbers when it is absolutely
necessary, and to instead use symbols to represent certain irrational numbers (such as it); approximating
them only at the very end of a calculation. If it is necessary to approximate a number in the middle of a
calculation, then it is often good enough to approximate to a few decimal places.

End of chapter exercises

1. Calculate:

a. yl6 — V72 to three decimal places

b. \/25 + \/2 to one decimal place

c. V48 — y/3 to two decimal places

d. \/64 + \/l8 — \/l2 to two decimal places

e. \/4 + \/20 — a/18 to six decimal places

f. \/3 + V5 — \/6 to one decimal place

2. Calculate:

a.

\Jx — 2, if x = 3, 3. Write the answer to four decimal places.

b. \/4 + x, if x = 1,423. Write the answer to two decimal places.

c. \Jx + 3 + yfx, if x = 5, 7. Write the answer to eight decimal places.

d. ^/2x — 5 + \\/x + 1, if x = 4, 91. Write the answer to five decimal places.

e. \J{ix — 1) + (4a; + 3) — \Jx + 5, if x = 3, 6. Write the answer to six decimal places.

f. f) \/(2x + 5) — (x — 1) + (5a; + 2) + |\/4 + x , if x = 1, 09. Write the answer to one decimal place

5

4 http:// www.fhsst.org/lbl
5 http:// www.fhsst.org/lb5

Introduction

In Grade 10, you learned about arithmetic sequences, where the difference between consecutive terms was

A quadratic sequence is a sequence of numbers in which the second differences between each
consecutive term differ by the same amount, called a common second difference.

For example,

1; 2; 4; 7; 11; ... (1)

is a quadratic sequence. Let us see why ...

If we take the difference between consecutive terms, then:

(2)

We then work out the second differences, which is simply obtained by taking the difference between the
consecutive differences {1; 2; 3; 4; ...} obtained above:

2- 1 = 1

3-2 = 1

(3)
4-3 = 1

We then see that the second differences are equal to 1. Thus, (1) is a quadratic sequence.

Note that the differences between consecutive terms (that is, the first differences) of a quadratic sequence
form a sequence where there is a constant difference between consecutive terms. In the above example, the
sequence of {1; 2; 3; 4; ...}, which is formed by taking the differences between consecutive terms of (1), has
a linear formula of the kind ax + b.

a 2 - ai

= 2-1

= 1

a 3 - a 2

= 4-2

= 2

a 4 - a 3

= 7-4

= 3

a§ — a 4

= 11-7

= 4

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14

The following are also examples of quadratic sequences:

3; 6

10; 15

21

4; 9

16; 25

36

7; 17

31; 49

71

2; 10

26; 50

82

31; 30

27; 22

15

(4)

Can you calculate the common second difference for each of the above examples?

Write down the next two terms and find a formula for the n th term of the sequence 5, 12,23,38, ..., ..

General Case

If the sequence is quadratic, the n th term should be T n = an 2 + bn + c

TERMS

a + b + c

4a + 2o+c

9a + 3o+c

1 st difference

3a +6

5a + b

7a +6

2 nd difference

2a

2a

Table 1

In each case, the second difference is 2a. This fact can be used to find a, then b then a

The following sequence is quadratic: 8, 22, 42, 68, ... Find the rule.

Derivation of the n th -term of a Quadratic Sequence

Let the n th -term for a quadratic sequence be given by

a n = A-n 2 + B-n+C
where A, B and C are some constants to be determined.

(5)

a n = A- n 2 + B ■ n + C

0l = A{lf + B {1) + C = A+B + C

a 2 = A{2) 2 + B{2) + C = 4:A+2B + C

a 3 = A{3f + B{3) + C=9A+3B + C

Letd =

a 2 - ax

:.d =

3A+ B

=>B =

d-3A

(6)

(7)
(8)

15

The common second difference

is obtained from

D

= (03 - a 2 ) -

(02 -

- ai)

= {5A+B)-

(3A + B)

2A

— !

Therefore,

from (8),

B = d---

2

L>

From (6),

C = oi-

-(A + B)=ai-

-< + §

•£>

(9)

(10)

(11)

(12)

.-. C = ai + .D-d (13)

Finally, the general equation for the n'^-term of a quadratic sequence is given by

a n = y -n 2 + U-^Dj -n+{ ai -d+D) (14)

Exercise 15: Using a set of equations

Study the following pattern: 1; 7; 19; 37; 61; ...

1. What is the next number in the sequence ?

2. Use variables to write an algebraic statement to generalise the pattern.

3. What will the 100th term of the sequence be ?

Plotting a graph of terms of a quadratic sequence

Plotting a n vs. n for a quadratic sequence yields a parabolic graph.

3; 6; 10; 15; 21; ... (15)

If we plot each of the terms vs. the corresponding index, we obtain a graph of a parabola.

Image not finished

Figure 1

16

End of chapter Exercises

1. Find the first 5 terms of the quadratic sequence defined by:

n + 2n + 1

(16)

2. Determine which of the following sequences is a quadratic sequence by calculating the common second
difference:

6; 9; 14; 21; 30;..

1; 7; 17; 31; 49;..
;80;
125
146
109

8; 17; 32; 53;

9; 26; 51; 84;

2; 20; 50; 92;

5; 19; 41; 71;
g. 2; 6; 10; 14; 18;
h. 3; 9; 15; 21; 27;
i. 10; 24; 44; 70; 102;...
j. 1; 2, 5; 5; 8,5; 13;...
k. 2, 5; 6; 10, 5; 16; 22,5;...
1. 0,5;9;20,5;35;52,5;...

2n 2 , find for which value of n, a r ,

2

242
■ (n — 4) , find for which value of n, a n = 36
n 2 + 4, find for which value of n, a n = 85
3n 2 , find an
7n 2 + An, find ag
An 2 + 3n — 1, find as
1, 5n 2 , find aio

10. For each of the quadratic sequences, find the common second difference, the formula for the general
term and then use the formula to find aioo-

4,7,12,19,28,...

2,8,18,32,50,...

7,13,23,37,55,...

5,14,29,50,77,...

7,22,47,82,127,...

3,10,21,36,55,...

3,7,13,21,31,...

3,9,17,27,39,...

3. Given a n

4. Given a n

5. Given a n

6. Given a n

7. Given a n

8. Given a n

9. Given a n

a.
b.
c.

d.

0.

f.

Chapter 1

Finance

1.1 Simple depreciation 1

1.1.1 Introduction

In Grade 10, the ideas of simple and compound interest were introduced. In this chapter we will be extending
those ideas, so it is a good idea to go back to the Finance chapter and revise what you learnt in Grade 10.
If you master the techniques in this chapter, you will understand about depreciation and will learn how to
determine which bank is offering the better interest rate.

1.1.2 Depreciation

It is said that when you drive a new car out of the dealership, it loses 20% of its value, because it is now
"second-hand". And from there on the value keeps falling, or depreciating. Second hand cars are cheaper
than new cars, and the older the car, usually the cheaper it is. If you buy a second hand (or should we say
pre-owned\) car from a dealership, they will base the price on something called book value.

The book value of the car is the value of the car taking into account the loss in value due to wear, age
and use. We call this loss in value depreciation, and in this section we will look at two ways of how this is
calculated. Just like interest rates, the two methods of calculating depreciation are simple and compound
methods.

The terminology used for simple depreciation is straight-line depreciation and for compound depreci-
ation is reducing-balance depreciation. In the straight-line method the value of the asset is reduced by
the same constant amount each year. In the compound depreciation method the value of the asset is reduced
by the same percentage each year. This means that the value of an asset does not decrease by a constant
amount each year, but the decrease is most in the first year, then by a smaller amount in the second year
and by an even smaller amount in the third year, and so on.

1.1.2.1 Depreciation

You may be wondering why we need to calculate depreciation. Determining the value of assets (as in the
example of the second hand cars) is one reason, but there is also a more financial reason for calculating
depreciation - tax! Companies can take depreciation into account as an expense, and thereby reduce their
taxable income. A lower taxable income means that the company will pay less income tax to the Revenue
Service.

1 This content is available online at <http://siyavula.cnx.Org/content/m38825/l.l/>.

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18

CHAPTER 1. FINANCE

1.1.3 Simple Depreciation (it really is simple!)

Let us go back to the second hand cars. One way of calculating a depreciation amount would be to assume
that the car has a limited useful life. Simple depreciation assumes that the value of the car decreases by an
equal amount each year. For example, let us say the limited useful life of a car is 5 years, and the cost of the
car today is R60 000. What we are saying is that after 5 years you will have to buy a new car, which means
that the old one will be valueless at that point in time. Therefore, the amount of depreciation is calculated:

R60 000

5 years

RU 000 per year.

The value of the car is then:

End of Year 1

R60 000- lx(R12 000)

= R48 000

End of Year 2

R60 000- 2x(R12 000)

= R36 000

End of Year 3

R60 000- 3x(R12 000)

= R24 000

End of Year 4

R60 000- 4x(R12 000)

= R12 000

End of Year 5

R60 000- 5x(R12 000)

= R0

Table 1.1

This looks similar to the formula for simple interest:

Total interest after n years = n X (P X i)

(1.1)

(1.2)

where i is the annual percentage interest rate and P is the principal amount.

If we replace the word interest with the word depreciation and the word principal with the words initial
value we can use the same formula:

Total depreciation after n years = n x [P x i)
Then the book value of the asset after n years is:

Initial value — Total depreciation after n years = P — n x [P x i)

= P(l — n x i)
For example, the book value of the car after two years can be simply calculated as follows:

(1.3)

(1.4)

Book value after 2 years

P(l-nx i)

Rm 000(1-2 x 20%)

7?60 000(1-0,4)

Rm 000 (0, 6)

#36 000

(1.5)

as expected.

Note that the difference between the simple interest calculations and the simple depreciation calculations
is that while the interest adds value to the principal amount, the depreciation amount reduces value!

Exercise 1.1: Simple Depreciation method (Solution on p. 29.)

A car is worth R240 000 now. If it depreciates at a rate of 15% p. a. on a straight-line depreciation,
what is it worth in 5 years' time ?

19

Exercise 1.2: Simple Depreciation (Solution on p. 29.)

A small business buys a photocopier for R 12 000. For the tax return the owner depreciates this
asset over 3 years using a straight-line depreciation method. What amount will he fill in on his tax
form after 1 year, after 2 years and then after 3 years ?

1.1.3.1 Salvage Value

Looking at the same example of our car with an initial value of R60 000, what if we suppose that we think we
would be able to sell the car at the end of the 5 year period for R10 000? We call this amount the "Salvage
Value"

We are still assuming simple depreciation over a useful life of 5 years, but now instead of depreciating
the full value of the asset, we will take into account the salvage value, and will only apply the depreciation
to the value of the asset that we expect not to recoup, i.e. R60 000 - R10 000 = R50 000.

The annual depreciation amount is then calculated as (R60 000 - R10 000) / 5 = R10 000

In general, the formula for simple (straight line) depreciation:

Initial value — Salvage value

Annual depreciation = (1-6)

Useful life

1.1.3.2 Simple Depreciation

1. A business buys a truck for R560 000. Over a period of 10 years the value of the truck depreciates to
R0 (using the straight-line method). What is the value of the truck after 8 years ?

2. Shrek wants to buy his grandpa's donkey for R800. His grandpa is quite pleased with the offer, seeing
that it only depreciated at a rate of 3% per year using the straight-line method. Grandpa bought the
donkey 5 years ago. What did grandpa pay for the donkey then ?

3. Seven years ago, Rocco's drum kit cost him R 12 500. It has now been valued at R2 300. What rate
of simple depreciation does this represent ?

4. Fiona buys a DsTV satellite dish for R3 000. Due to weathering, its value depreciates simply at 15%
per annum. After how long will the satellite dish be worth nothing ?

1.2 Compound depreciation 2
1,2,1 Compound Depreciation

The second method of calculating depreciation is to assume that the value of the asset decreases at a certain
annual rate, but that the initial value of the asset this year, is the book value of the asset at the end of last
year.

For example, if our second hand car has a limited useful life of 5 years and it has an initial value of
R60 000, then the interest rate of depreciation is 20% (100%/5 years). After 1 year, the car is worth:

Book value after first year = P (1 — n X i)

= i?60 000(1- 1 x 20%)

#60 000(1-0,2) (1.7)

R60 000(0,8)
i?48 000

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CHAPTER 1. FINANCE

At the beginning of the second year, the car is now worth R48 000, so after two years, the car is worth:

Book value after second year = P (1 — n x i)

= #48 000(1-1x20%)
i?48 000(1-0,2)
i?48 000(0,8)
R38 400
We can tabulate these values.

(1.8)

End of first year

R60 000(1 - 1 x 20%)=R60 000(1 - 1 x 20%) 1

= R48 000,00

End of second year

R48 000 (1 - 1 x 20%)=R60 000(1 - 1 x 20%) 2

= R38 400,00

End of third year

R38 400(1 - 1 x 20%)=R60 000(1 - 1 x 20%) 3

= R30 720,00

End of fourth year

R30 720(1 - 1 x 20%)=R60 000(1 - 1 x 20%) 4

= R24 576,00

End of fifth year

R24 576(1 - 1 x 20%)=R60 000(1 - 1 x 20%) 5

= R19 608,80

Table 1.2

We can now write a general formula for the book value of an asset if the depreciation is compounded.

Initial value — Total depreciation after n years = P(l — i) (1-9)

For example, the book value of the car after two years can be simply calculated as follows:

Book value after 2 years

P(l-i) n

R60 000(1 - 20%) 2

R60 000(1 -0,2) 2

R60 000(0, 8) 2

R38 400

(1.10)

as expected.

Note that the difference between the compound interest calculations and the compound depreciation
calculations is that while the interest adds value to the principal amount, the depreciation amount reduces
value!

Exercise 1.3: Compound Depreciation (Solution on p. 29.)

The Flamingo population of the Bergriver mouth is depreciating on a reducing balance at a rate
of 12% p. a. If there are now 3 200 flamingos in the wetlands of the Bergriver mouth, how many
will there be in 5 years' time ? Answer to three significant numbers.

Exercise 1.4: Compound Depreciation (Solution on p. 29.)

Farmer Brown buys a tractor for R250 000 and depreciates it by 20% per year using the compound
depreciation method. What is the depreciated value of the tractor after 5 years ?

21

1.2.1.1 Compound Depreciation

1. On January 1, 2008 the value of my Kia Sorento is R320 000. Each year after that, the cars value will
decrease 20% of the previous year [U+0092] s value. What is the value of the car on January 1, 2012?

2. The population of Bonduel decreases at a rate of 9,5% per annum as people migrate to the cities.
Calculate the decrease in population over a period of 5 years if the initial population was 2 178 000.

3. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses 3% of its water each
day. How much does it weigh after a month of 31 days ?

4. A computer depreciates at x% per annum using the reducing-balance method. Four years ago the
value of the computer was R10 000 and is now worth R4 520. Calculate the value of x correct to two
decimal places.

1.3 Present and future values 3

1.3.1 Present Values or Future Values of an Investment or Loan

1.3.1.1 Now or Later

When we studied simple and compound interest we looked at having a sum of money now, and calculating
what it will be worth in the future. Whether the money was borrowed or invested, the calculations examined
what the total money would be at some future date. We call these future values.

It is also possible, however, to look at a sum of money in the future, and work out what it is worth now.
This is called a present value.

For example, if Rl 000 is deposited into a bank account now, the future value is what that amount will
accrue to by some specified future date. However, if Rl 000 is needed at some future time, then the present
value can be found by working backwards - in other words, how much must be invested to ensure the money
grows to Rl 000 at that future date?

The equation we have been using so far in compound interest, which relates the open balance (P), the
closing balance (A), the interest rate (i as a rate per annum) and the term (n in years) is:

A = P-(l + i) n (1.11)

Using simple algebra, we can solve for P instead of A, and come up with:

p = A-{i + iy n (i.i2)

This can also be written as follows, but the first approach is usually preferred.

P = i vn (1-13)

(1 + t)

Now think about what is happening here. In Equation (1.11), we start off with a sum of money and we let
it grow for n years. In Equation (1.12) we have a sum of money which we know in n years time, and we
"unwind" the interest - in other words we take off interest for n years, until we see what it is worth right
now.

We can test this as follows. If I have Rl 000 now and I invest it at 10% for 5 years, I will have:

A = P-(l + i) n

= Rl 000(1 + 10%) 5 (1.14)

PI 610,51

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22 CHAPTER 1. FINANCE

at the end. BUT, if I know I have to have Rl 610,51 in 5 years time, I need to invest:

p = A-(l + i)' n

= i?l 610,51(1 + 10%) -5 (1.15)

Rl 000

We end up with Rl 000 which - if you think about it for a moment - is what we started off with. Do you
see that?

Of course we could apply the same techniques to calculate a present value amount under simple interest
rate assumptions - we just need to solve for the opening balance using the equations for simple interest.

A = P(l + ixn) (1.16)

Solving for P gives:

P = A/(l + ixn) (1.17)

Let us say you need to accumulate an amount of Rl 210 in 3 years time, and a bank account pays Simple
Interest of 7%. How much would you need to invest in this bank account today?

P =

1+n-i

_ m 2io a 18 \

- 1+3x7% l 1 - 1 ^

= Rl 000

Does this look familiar? Look back to the simple interest worked example in Grade 10. There we started
with an amount of Rl 000 and looked at what it would grow to in 3 years' time using simple interest rates.
Now we have worked backwards to see what amount we need as an opening balance in order to achieve the
closing balance of Rl 210.

In practice, however, present values are usually always calculated assuming compound interest. So unless
you are explicitly asked to calculate a present value (or opening balance) using simple interest rates, make
sure you use the compound interest rate formula!

1.3.1.1.1 Present and Future Values

1. After a 20- year period Josh's lump sum investment matures to an amount of R313 550. How much
did he invest if his money earned interest at a rate of 13,65% p. a. compounded half yearly for the first
10 years, 8,4% p. a. compounded quarterly for the next five years and 7,2% p. a. compounded monthly
for the remaining period ?

2. A loan has to be returned in two equal semi-annual instalments. If the rate of interest is 16% per
annum, compounded semi-annually and each instalment is Rl 458, find the sum borrowed.

1.4 Finding i and n 4

1.4.1 Finding i

By this stage in your studies of the mathematics of finance, you have always known what interest rate to
use in the calculations, and how long the investment or loan will last. You have then either taken a known
starting point and calculated a future value, or taken a known future value and calculated a present value.
But here are other questions you might ask:

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23

1. I want to borrow R2 500 from my neighbour, who said I could pay back R3 000 in 8 months time.
What interest is she charging me?

2. I will need R450 for some university textbooks in 1,5 years time. I currently have R400. What interest
rate do I need to earn to meet this goal?

Each time that you see something different from what you have seen before, start off with the basic equation
that you should recognise very well:

A = P-{l + t) n (1.19)

If this were an algebra problem, and you were told to "solve for i", you should be able to show that:

I = (1 + 0"
(1 + i) = (^) lAl (1-20)

i = tf) 1 '"-!

You do not need to memorise this equation, it is easy to derive any time you need it!
So let us look at the two examples mentioned above.

1. Check that you agree that P=R2 500, A=R3 000, n=8/12=0,666667. This means that:

• _ ( A3 poo \ 1/0,666667

' — \R2 500 i i

31,45%

Ouch! That is not a very generous neighbour you have.

2. Check that P=R400, .4=R450, n=l,5

/ il450 \ 1 / 1 ' 5 _ i

1 ~ V_R400^ i

8, 17%

(1.21)

(1.22)

This means that as long as you can find a bank which pays more than 8,17% interest, you should have
the money you need!

Note that in both examples, we expressed n as a number of years (y| years, not 8 because that is the number
of months) which means % is the annual interest rate. Always keep this in mind - keep years with years to
avoid making silly mistakes.

1.4.1.1 Finding i

1. A machine costs R45 000 and has a scrap value of R9 000 after 10 years. Determine the annual rate
of depreciation if it is calculated on the reducing balance method.

2. After 5 years an investment doubled in value. At what annual rate was interest compounded ?

1.4.2 Finding n - Trial and Error

By this stage you should be seeing a pattern. We have our standard formula, which has a number of variables:

A = P-(l + i) n (1.23)

We have solved for A (in Grade 10), P (in "Present Values or Future Values of an Investment or Loan")
and i (in "Finding i" (Section 1.4.1: Finding «)). This time we are going to solve for n. In other words, if

24 CHAPTER 1. FINANCE

we know what the starting sum of money is and what it grows to, and if we know what interest rate applies
- then we can work out how long the money needs to be invested for all those other numbers to tie up.

This section will calculate n by trial and error and by using a calculator. The proper algebraic solution
will be learnt in Grade 12.

Solving for n, we can write:

P(l

P = (! + *)

(1.24)

Now we have to examine the numbers involved to try to determine what a possible value of n is. Refer to

Exercise 1.5: Term of Investment - Trial and Error (Solution on p. 30.)

We invest R3 500 into a savings account which pays 7,5% compound interest for an unknown
period of time, at the end of which our account is worth R4 044,69. How long did we invest the
money?

1.4.2.1 Finding n - Trial and Error

1. A company buys two types of motor cars: The Acura costs R80 600 and the Brata R101 700 VAT
included. The Acura depreciates at a rate, compounded annually, of 15,3% per year and the Brata
at 19,7%, also compounded annually, per year. After how many years will the book value of the two
models be the same ?

2. The fuel in the tank of a truck decreases every minute by 5,5% of the amount in the tank at that point
in time. Calculate after how many minutes there will be less than 30/ in the tank if it originally held
200L

1.5 Nominal and effective interest rates 5
1.5.1 Nominal and Effective Interest Rates

So far we have discussed annual interest rates, where the interest is quoted as a per annum amount. Although
it has not been explicitly stated, we have assumed that when the interest is quoted as a per annum amount
it means that the interest is paid once a year.

Interest however, may be paid more than just once a year, for example we could receive interest on a
monthly basis, i.e. 12 times per year. So how do we compare a monthly interest rate, say, to an annual
interest rate? This brings us to the concept of the effective annual interest rate.

One way to compare different rates and methods of interest payments would be to compare the Closing
Balances under the different options, for a given Opening Balance. Another, more widely used, way is
to calculate and compare the "effective annual interest rate" on each option. This way, regardless of the
differences in how frequently the interest is paid, we can compare apples-with-apples.

For example, a savings account with an opening balance of Rl 000 offers a compound interest rate of 1%
per month which is paid at the end of every month. We can calculate the accumulated balance at the end
of the year using the formulae from the previous section. But be careful as our interest rate has been given
as a monthly rate, so we need to use the same units (months) for our time period of measurement.

tip: Remember, the trick to using the formulae is to define the time period, and use the interest
rate relevant to the time period.

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25

So we can calculate the amount that would be accumulated by the end of 1-year as follows:

Closing balance after 12 months = P X (1 + i)

= Rl 000 x (1 + 1%) 12 (1.25)

Rl 126,83

Note that because we are using a monthly time period, we have used n = 12 months to calculate the balance
at the end of one year.

The effective annual interest rate is an annual interest rate which represents the equivalent per annum
interest rate assuming compounding.

It is the annual interest rate in our Compound Interest equation that equates to the same accumulated
balance after one year. So we need to solve for the effective annual interest rate so that the accumulated
balance is equal to our calculated amount of Rl 126,83.

We use i 12 to denote the monthly interest rate. We have introduced this notation here to distinguish
between the annual interest rate, i. Specifically, we need to solve for i in the following equation:

Pxfl + j) 1 = Px(l + i 12 ) 12

(1-H) = (l + «i 2 ) 12 divide both sides by P (1.26)

i = (1 + i 12 ) — 1 subtract 1 from both sides

For the example, this means that the effective annual rate for a monthly rate «i2 = 1% is:

* = (l + *i 2 ) 12 -l
= (1 + 1%) 12 -1
0,12683
12, 683%
If we recalculate the closing balance using this annual rate we get:

Closing balance after 1 year = Px(lfi)

= m ooo x (1 + 12, 683%) 1 (1.28)

Rl 126,83

which is the same as the answer obtained for 12 months.

Note that this is greater than simply multiplying the monthly rate by 12 (12 x 1% = 12%) due to the
effects of compounding. The difference is due to interest on interest. We have seen this before, but it is an
important point!

1.5.1.1 The General Formula

So we know how to convert a monthly interest rate into an effective annual interest. Similarly, we can convert
a quarterly interest, or a semi-annual interest rate or an interest rate of any frequency for that matter into
an effective annual interest rate.

For a quarterly interest rate of say 3% per quarter, the interest will be paid four times per year (every
three months) . We can calculate the effective annual interest rate by solving for i:

P(1-H) = P(1-H 4 ) 4 (1.29)

where i^ is the quarterly interest rate.

So (1 + 1) = (1, 03) , and so i = 12, 55%. This is the effective annual interest rate.

(1.27)

26 CHAPTER 1. FINANCE

In general, for interest paid at a frequency of T times per annum, the follow equation holds:

P(l + i) = P(l + i T ) T (1.30)

where it is the interest rate paid T times per annum.

1.5.1.2 De-coding the Terminology

Market convention however, is not to state the interest rate as say 1% per month, but rather to express this
amount as an annual amount which in this example would be paid monthly. This annual amount is called
the nominal amount.

The market convention is to quote a nominal interest rate of "12% per annum paid monthly" instead of
saying (an effective) 1% per month. We know from a previous example, that a nominal interest rate of 12%
per annum paid monthly, equates to an effective annual interest rate of 12,68%, and the difference is due to
the effects of interest-on- interest.

So if you are given an interest rate expressed as an annual rate but paid more frequently than annual, we
first need to calculate the actual interest paid per period in order to calculate the effective annual interest
rate.

Nominal interest rate per annum

monthly interest rate = : (1-31)

number of periods per year

For example, the monthly interest rate on 12% interest per annum paid monthly, is:

monthly interest rate = Nominal interest rate per annum
J number ot periods per year

12% (-1 on)

- 12 months K 1 -^)

= 1% per month

The same principle applies to other frequencies of payment.

Exercise 1.6: Nominal Interest Rate (Solution on p. 30.)

Consider a savings account which pays a nominal interest at 8% per annum, paid quarterly.
Calculate (a) the interest amount that is paid each quarter, and (b) the effective annual interest
rate.

Exercise 1.7: Nominal Interest Rate (Solution on p. 31.)

On their saving accounts, Echo Bank offers an interest rate of 18% nominal, paid monthly. If you
save R100 in such an account now, how much would the amount have accumulated to in 3 years'
time?

1.5.1.2.1 Nominal and Effect Interest Rates

1. Calculate the effective rate equivalent to a nominal interest rate of 8,75% p. a. compounded monthly.

2. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every four months on her
investment of R 85 000. Calculate the effective rate per annum.

1,5.2 Formulae Sheet

As an easy reference, here are the key formulae that we derived and used during this chapter. While
memorising them is nice (there are not many), it is the application that is useful. Financial experts are not
paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve financial
problems.

27

1.5.2.1 Definitions

P

Principal (the amount of money at the starting point of the calculation)

interest rate, normally the effective rate per annum

period for which the investment is made

ix the interest rate paid T times per annum, i.e. ij-

Nominal interest rate

Table 1.3

1.5.2.2 Equations

Simple increase : A = P (I + i x n)

Compound increase : A = P(\ + i)

Simple decrease : A = P (I — i x n)

Compound decrease : A = P(l — i)

Effective annual interest rate (i) : (1 + i) = (1 + it)

(1.33)

1,5.3 End of Chapter Exercises

1. Shrek buys a Mercedes worth R385 000 in 2007. What will the value of the Mercedes be at the end of
2013 if

a. the car depreciates at 6% p. a. straight-line depreciation

b. the car depreciates at 12% p. a. reducing-balance depreciation.

2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R8 900. The interest rate is
quoted as 11% per annum based on simple interest. Calculate the required monthly payment for this
contract.

3. A computer is purchased for R16 000. It depreciates at 15% per annum.

a. Determine the book value of the computer after 3 years if depreciation is calculated according to
the straight-line method.

b. Find the rate, according to the reducing-balance method, that would yield the same book value
as in list, p. 27 after 3 years.

4. Maggie invests R12 500,00 for 5 years at 12% per annum compounded monthly for the first 2 years
and 14% per annum compounded semi-annually for the next 3 years. How much will Maggie receive
in total after 5 years?

5. Tintin invests R120 000. He is quoted a nominal interest rate of 7,2% per annum compounded monthly.

a. Calculate the effective rate per annum correct to THREE decimal places.

b. Use the effective rate to calculate the value of Tintin's investment if he invested the money for 3
years.

c. Suppose Tintin invests his money for a total period of 4 years, but after 18 months makes a
withdrawal of R20 000, how much will he receive at the end of the 4 years?

6. Paris opens accounts at a number of clothing stores and spends freely. She gets heself into terrible
debt and she cannot pay off her accounts. She owes Hilton Fashion world R5 000 and the shop agrees
to let Paris pay the bill at a nominal interest rate of 24% compounded monthly.

28 CHAPTER 1. FINANCE

a. How much money will she owe Hilton Fashion World after two years ?

b. What is the effective rate of interest that Hilton Fashion World is charging her ?

29

Solutions to Exercises in Chapter 1

Solution to Exercise 1.1 (p. 18)

Step 1.

p

=

7?240 000

i

=

0,15

n

=

5

(1.34)

A is required

Step 2.

A = 240 000(1-0,15x5) (1.35)

Step 3.

240 000(1-0,75)
240 000x0,25 (1-36)

60 000
Step 4. In 5 years' time the car is worth R60 000
Solution to Exercise 1.2 (p. 19)

Step 1. The owner of the business wants the photocopier to depreciate to R0 after 3 years. Thus, the value of

the photocopier will go down by 12 000 -r 3 = RA 000 per year.
Step 2. 12 000 - 4 000 = R8 000
Step 3. 8 000 - 4 000 = RA 000
Step 4. 4 000 - 4 000 =

After 3 years the photocopier is worth nothing

Solution to Exercise 1.3 (p. 20)

Step 1.

P = R3 200

i = 0,12

n = 5

A is required

Step 2.

A = 3 200(1 -0,12) 5

(1.37)

(1.38)

Step 3.

A = 3 200(0, 88) 5

= 3 200x0,527731916 (1.39)

1688,742134
Step 4. There would be approximately 1 690 flamingos in 5 years' time.
Solution to Exercise 1.4 (p. 20)

30

CHAPTER 1. FINANCE

Step 1.

P

i

n

R250 000

0,2

5

Step 2.
Step 3.

A is required

= 250 000(1 -0,2) 5

250 000(0, 8) 5
= 250 000 x 0, 32768
81 920

Step 4. Depreciated value after 5 years is R 81 920
Solution to Exercise 1.5 (p. 24)

Step 1. • P=R3 500

• i=7,5%

• A=R4 044,69

We are required to find n.
Step 2. We know that:

Step 3.

A =

P(l + i) n

A

P ~

(l + i) n

Ri 044,69
R3 500

= (1 + 7,5%)"

1,156

(1,075)"

(1.40)

(1.41)

(1.42)

(1.43)

(1.44)

We now use our calculator and try a few values for n.

Possible n

1,075"

1,0

1,075

1,5

1,115

2,0

1,156

2,5

1,198

Table 1.4

We see that n is close to 2.
Step 4. The R3 500 was invested for about 2 years.

Solution to Exercise 1.6 (p. 26)

Step 1. We are given that a savings account has a nominal interest rate of ,
to find:

paid quarterly. We are required

the quarterly interest rate, i^

31

• the effective annual interest rate, i
Step 2. We know that:

and

Nominal interest rate per annum

quarterly interest rate = (1-45)

number of quarters per year

P (1 + i) = p(i + i T f (1.46)

where T is 4 because there are 4 payments each year.
Step 3.

, i ■ , , , Nominal interest rate per annum

quarterly interest rate = r -? = — t—

^ J number or periods per year

_ 8%

4 quarters

2% per quarter

(1.47)

Step 4. The effective annual interest rate (i) is calculated as:

(1 + i) = (l + z 4 ) 4

(1 + i) = (l + 2%) 4
i = (l + 2%) 4 -l
8, 24%

(1.48)

Step 5. The quarterly interest rate is 2% and the effective annual interest rate is 8,24%, for a nominal interest
rate of 8% paid quarterly.

Solution to Exercise 1.7 (p. 26)

Step 1. Interest rate is 18% nominal paid monthly. There are 12 months in a year. We are working with
a yearly time period, so n = 3. The amount we have saved is R100, so P = 100. We need the
accumulated value, A.

Step 2. We know that

Nominal interest rate per annum

monthly interest rate = : (1-49)

number of periods per year

for converting from nominal interest rate to effective interest rate, we have

l + i=(l + i T ) T (1.50)

and for calculating accumulated value, we have

A = Px(l + i) n (1.51)

Step 3. There are 12 month in a year, so

Nominal annual interest rate

*12 - 12

_ 18%

12

(1.52)

1, 5% per month

32 CHAPTER 1. FINANCE

and then, we have

(1.53)

1 + i =

(l + ii2) 12

i =

(l + i 12 ) 12 -l

=

(1 + 1,5%) 12 -1

=

(1,015) 12 - 1

=

19, 56%

A =

Px (l + if

,n w /i i in ccW \ 3

Step 4.

100x1,7091
170,91

Step 5. The accumulated value is R170, 91. (Remember to round off to the the nearest cent.)

(1.54)

Chapter 2

2.1 Factorisation 1

2.1.1 Introduction

In grade 10, the basics of solving linear equations, quadratic equations, exponential equations and linear
inequalities were studied. This chapter extends on that work. We look at different methods of solving

2.1.2 Solution by Factorisation

How to solve quadratic equations by factorisation was discussed in Grade 10. Here is an example to remind
you of what is involved.

Exercise 2.1: Solution of Quadratic Equations (Solution on p. 43.)

Solve the equation 2a; 2 — 5x — 12 = 0.

It is important to remember that a quadratic equation has to be in the form ax 2 + bx + c= before one can
solve it using the following methods.

Exercise 2.2: Solving quadratic equation by factorisation (Solution on p. 43.)

Solve for a: a (a — 3) = 10

Exercise 2.3: Solving fractions that lead to a quadratic equation (Solution on p. 44.)
Solve for b: *L + i = _4_

2.1.2.1 Solution by Factorisation

Solve the following quadratic equations by factorisation. Some answers may be left in surd form.

1. 2y 2 -61 = 101

2. 2y 2 - 10 =

3. y 2 - 4 = 10

4. 2y 2 - 8 = 28

5. 7y 2 = 28

6. y 2 + 28 = 100

7. 7y 2 + Uy =

8. I2y 2 + 24y + 12 =

9. 16y 2 - 400 =

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33

34 CHAPTER 2. SOLVING QUADRATIC EQUATIONS

10. y 2 -5y + 6 =

11. y 2 + 5y-36 =

12. y 2 + 2y = 8

13. -y 2 - lly-2A =

14. 13y - 42 = y 2

15. y 2 + 9y+ 14 =

16. y 2 -5ky + 4k 2 =

17. y(2y+l) = 15

18. J1L+ 3 +2 - _=«_

j/-2 j/ y 2 -2y

S/-2 _ 2y+l
J/+1 y-7

2.2 Completing the square 2

2,2,1 Solution by Completing the Square

We have seen that expressions of the form:

a 2 x 2 - b 2 (2.1)

are known as differences of squares and can be factorised as follows:

(ax - b) [ax + b) . (2.2)

This simple factorisation leads to another technique to solve quadratic equations known as completing the
square.

We demonstrate with a simple example, by trying to solve for x in:

x 2 - 2x - 1 = 0. (2.3)

We cannot easily find factors of this term, but the first two terms look similar to the first two terms of the
perfect square:

(a; - l) 2 = x 2 - 2x + 1. (2.4)

However, we can cheat and create a perfect square by adding 2 to both sides of the equation in (2.3) as:

(2.5)

Now we know that:

which means that:

x 2 - 2x - 1 =

x 2 - 2x - 1 + 2 =

+ 2

x 2 - 2x + 1 =

2

(x-lf =

2

(x-lf -2 =

2=(V2) 2

(x-lf -2

(2.6)

(2.7)

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35

is a difference of squares. Therefore we can write:

(x - l) 2 - 2 = \(x - 1) - \/2~l [(a; - 1) + \/2~l = 0. (2.i

The solution to x 2 — 2x — 1 = is then:

(a; - 1) - \/2 = (2.9)

(a; - 1) + V2 = 0. (2.10)

This means x = 1 + -\/2 or i = 1 - \/2- This example demonstrates the use of completing the square to

2.2.1.1 Method: Solving Quadratic Equations by Completing the Square

1. Write the equation in the form ax 2 + bx + c = 0. e.g. x 2 + 2x — 3 =

2. Take the constant over to the right hand side of the equation, e.g. x 2 + 2x = 3

3. If necessary, make the coefficient of the x 2 term = 1, by dividing through by the existing coefficient.

4. Take half the coefficient of the x term, square it and add it to both sides of the equation, e.g. in
x 2 + 2x = 3, half of the coefficient of the x term is 1 and l 2 = 1. Therefore we add 1 to both sides to
get: x 2 + 2x+ 1 = 3+ 1.

5. Write the left hand side as a perfect square: (x + 1) —4 =

6. You should then be able to factorise the equation in terms of difference of squares and then solve for
x: (x+ l-2)(x + l + 2) =

Exercise 2.4: Solving Quadratic Equations by Completing the Square (Solution on p.

44.)

Solve:

x 2 - lOx- 11 = (2.11)

by completing the square

Exercise 2.5: Solving Quadratic Equations by Completing the Square (Solution on p.

45.)

Solve:

2a; 2 - 8a; - 16 = (2.12)

by completing the square

Figure 2.1

36 CHAPTER 2. SOLVING QUADRATIC EQUATIONS

2.2.1.1.1 Solution by Completing the Square

Solve the following equations by completing the square:

1. x 2 + lOz-2 =

2. x 2 +4a; + 3 =

3. x 2 + 8x - 5 =

4. 2s 2 + 12x + 4 =

5. x 2 + 5x + 9 =

6. x 2 + 16a; + 10 =

7. 3a; 2 + 6a; - 2 =

8. z 2 + 8z - 6 =

9. 2z 2 - llz =
10. 5 + 4z - z 2 =

2,3.1 Solution by the Quadratic Formula

It is not always possible to solve a quadratic equation by factorising and sometimes it is lengthy and tedious
to solve a quadratic equation by completing the square. In these situations, you can use the quadratic
formula that gives the solutions to any quadratic equation.
Consider the general form of the quadratic function:

/ (x) = ax 2 + bx + c. (2.13)

Factor out the a to get:

fix) = a(x 2 + -x+ - ) . (2.14)

\aaj

Now we need to do some detective work to figure out how to turn (2.14) into a perfect square plus some
extra terms. We know that for a perfect square:

(in + n) = m + 2mn + n (2-15)

and

(in — n) = in — 2mn + n (2-16)

The key is the middle term on the right hand side, which is 2x the first term x the second term of the
left hand side. In (2.14), we know that the first term is x so 2x the second term is -. This means that the
second term is w-. So,

7\2 7 / 1 \ 2

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37

In general if you add a quantity and subtract the same quantity, nothing has changed. This means if we
add and subtract (^-) from the right hand side of (2.14) we will get:

f{x) = a{x 2 + b -x+^)

= «(- 2 + !- + (^) 2 -(^) 2 + f)
«(> + (^)] 2 )

We set / (x) = to find its roots, which yields:

(2.18)

C 4a

bV b 2

a \ X+ Ya =Ta~ C (2 - 19)

Now dividing by a and taking the square root of both sides gives the expression

b / b 2 c ,

Finally, solving for x implies that

_b_ i / b 2
' 2a ZE \J Aa 2

b i / b 2 — 4oc

' 2a ^ V 4a 2

which can be further simplified to:

-6± V& 2 -4c

(2.21)

(2.22)

2a

These are the solutions to the quadratic equation. Notice that there are two solutions in general, but these
may not always exists (depending on the sign of the expression b 2 — Aac under the square root). These
solutions are also called the roots of the quadratic equation.

Exercise 2.6: Using the quadratic formula (Solution on p. 45.)

Find the roots of the function / (x) = 2x 2 + 3x — 7.

Exercise 2.7: Using the quadratic formula but no solution (Solution on p. 46.)

Find the solutions to the quadratic equation x 2 — 5x + 8 = 0.

Figure 2.2

38

2.3.1.1 Solution by the Quadratic Formula

Solve for t using the quadratic formula.

1. St 2

t

2. t 2 -M + 9 =

3. 2t 2 + 6t+5 =

4. 4£ 2 + 2£+2 =

5. -St 2 + 5t - 8 =

6. -ht 2 + St - 3 =

7. i 2 - At + 2 =

8. 9t 2 - It - 9 =

9. 2t 2 + 3i+2 =
10. t 2 + t+l =

TIP:

• In all the examples done so far, the solutions were left in surd form. Answers can also be
given in decimal form, using the calculator. Read the instructions when answering questions
in a test or exam whether to leave answers in surd form, or in decimal form to an appropriate
number of decimal places.

• Completing the square as a method to solve a quadratic equation is only done when specifically

2.3.1.2 Mixed Exercises

Solve the quadratic equations by either factorisation, completing the square or by using the quadratic formula:

• Always try to factorise first, then use the formula if the trinomial cannot be factorised.

• Do some of them by completing the square and then compare answers to those done using the other

methods.

1. 24y 2 + 61y-8 =

2. -8y 2 - 16y + 42 =

3. -9y 2 + 2Ay - 12 =

4. -5y 2 + 0y + 5 =

5. -3y 2 + 15y- 12 =

6. 49y 2 + Oy - 25 =

7. -12y 2 + 66y - 72 =

8. -40y 2 + 58y- 12 =

9. -24y 2 + 37y + 72 =

10. 6y 2 + ly - 24 =

11. 2y 2 -5y-3 =

12. -18j/ 2 -55y-25 =

13. -25y 2 + 25y - 4 =

14. -32y 2 + 24y + 8 =

15. 9y 2 - 13y- 10 =

16. 35y 2 - 8y - 3 =

17. -81y 2 -99y- 18 =

18. 14y 2 -81y+81 =

19. -Ay 2 - A\y - 45 =

20. 16y 2 + 20y - 36 =

21. 42y 2 + 104y+64 =

22. 9y 2 - 76y + 32 =

23. -54y 2 + 21y + 3 =

24. 36y 2 + AAy + 8 =

25. 64y 2 + 96y + 36 =

26. 12y 2 - 22y - 14 =

27. 16y 2 + Qy - 81 =

28. 3y 2 + lOy - 48 =

29. -Ay 2 + 8y - 3 =

30. -5y 2 - 2Qy + 63 =

31. x 2 - 70 = 11

32. 2a; 2 - 30 = 2

33. x 2 - 16 = 2 - x 2

34. 2y 2 - 98 =

35. 5y 2 - 10 = 115

36. 5y 2 - 5 = 19 - y 2

Table 2.1

39

2.4 Finding the equation 4

2.4.1 Finding an equation when you know its roots

We have mentioned before that the roots of a quadratic equation are the solutions or answers you get from
solving the quadatic equation. Working back from the answers, will take you to an equation.

Exercise 2.8: Find an equation when roots are given (Solution on p. 46.)

Find an equation with roots 13 and -5

Exercise 2.9: Fraction roots (Solution on p. 46.)

Find an equation with roots — | and 4

This section is not in the syllabus, but it gives one a good understanding about some of the solutions of the

2.4.1.1.1 What is the Discriminant of a Quadratic Equation?

Consider a general quadratic function of the form / (x) = ax 2 + bx + c. The discriminant is defined as:

A = b 2 - 4ac. (2.23)

This is the expression under the square root in the formula for the roots of this function. We have already
seen that whether the roots exist or not depends on whether this factor A is negative or positive.

2.4.1.1.2 The Nature of the Roots

2.4.1.1.2.1 Real Roots (A > 0)

Consider A > for some quadratic function / (x) = ax 2 + bx + c. In this case there are solutions to the
equation / (x) = given by the formula

-b ± Vb 2 - Aac -b±VA , A .

(2.24)

2a 2a

If the expression under the square root is non-negative then the square root exists. These are the roots of
the function / (x).

There various possibilities are summarised in the figure below.

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40 CHAPTER 2. SOLVING QUADRATIC EQUATIONS

I I

A ■\$ - imaginary roots A > - rea I loots

I ' 1

A=-0 A =

unequal root equal root

, I

A 3 perfect A not j per-

aqujre- ratio- feet square

njl roots - irrational

root

Figure 2.3

41

2.4.1.1.2.2 Equal Roots (A = 0)

If A = 0, then the roots are equal and, from the formula, these are given by

b

(2 - 25)

2.4.1.1.2.3 Unequal Roots (A > 0)

There will be 2 unequal roots if A > 0. The roots of / (x) are rational if A is a perfect square (a number
which is the square of a rational number), since, in this case, y/~K is rational. Otherwise, if A is not a perfect
square, then the roots are irrational.

2.4.1.1.2.4 Imaginary Roots (A < 0)

If A < 0, then the solution to / (x) = ax 2 + bx + c = contains the square root of a negative number and
therefore there are no real solutions. We therefore say that the roots of / (x) are imaginary (the graph of
the function / (x) does not intersect the x-axis).

Figure 2.4

2.4.1.1.2.5.1 From past papers

IEB, Nov. 2001, HG Given: x 2 + bx - 2 + k (x 2 + 3x + 2) = (k ^ -1)

a. Show that the discriminant is given by:

A = k 2 + Qbk + b 2 + 8 (2.26)

b. If b = 0, discuss the nature of the roots of the equation.

c. If b = 2, find the value(s) of k for which the roots are equal.

IEB, Nov. 2002, HG Show that k 2 x 2 + 2 = kx — x 2 has non-real roots for all real values for k.
IEB, Nov. 2003, HG The equation x 2 + 12x = 3fcx 2 + 2 has real roots.

a. Find the largest integral value of k.

b. Find one rational value of k, for which the above equation has rational roots.

IEB, Nov. 2003, HG In the quadratic equation px 2 + qx + r = 0, p, q and r are positive real numbers and form a geometric

sequence. Discuss the nature of the roots.

IEB, Nov. 2004, HG Consider the equation:

x 2 — 4 5

k = where i^- (2.27)

2x - 5 r 2 v '

42 CHAPTER 2. SOLVING QUADRATIC EQUATIONS

a. Find a value of k for which the roots are equal.

b. Find an integer k for which the roots of the equation will be rational and unequal.

IEB, Nov. 2005, HG a. Prove that the roots of the equation x 2 — (a + b) x + ab — p 2 = are real for all real values of a,

b and p.
b. When will the roots of the equation be equal?
IEB, Nov. 2005, HG If b and c can take on only the values 1, 2 or 3, determine all pairs (&; c) such that x 2 + bx + c = has
real roots.

2,4,2 End of Chapter Exercises

1. Solve: x 2 — x — 1 = (Give your answer correct to two decimal places.)

2. Solve: 16 (x + 1) = x 2 (x + 1)

3. Solve: y 2 + 3 H — f^g = 7 (Hint: Let y 2 + 3 = k and solve for k first and use the answer to solve y.)

4. Solve for x: 2x 4 - 5x 2 - 12 =

5. Solve for x:

a. x(x- 9) + 14 =

b. x 2 — x = 3 (Show your answer correct to ONE decimal place.)

c. x + 2 = - (correct to 2 decimal places)
d- -tt + ^K = 1

6. Solve for x by completing the square: x — px — 4 =

2
.3

7. The equation ax 2 + bx + c = has roots x = | and x = —4. Find one set of possible values for a, b

and c.

8. The two roots of the equation 4a; 2 + px — 9 = differ by 5. Calculate the value of p.

9. An equation of the form x 2 + bx + c = is written on the board. Saskia and Sven copy it down
incorrectly. Saskia has a mistake in the constant term and obtains the solutions -4 and 2. Sven has
a mistake in the coefficient of x and obtains the solutions 1 and -15. Determine the correct equation
that was on the board.

10. Bjorn stumbled across the following formula to solve the quadratic equation ax 2 + bx + c = in a

foreign textbook.

2c

x = ; 2.28

-b±Vb 2 -4ac

a. Use this formula to solve the equation:

2a; 2 + a;-3 = (2.29)

b. Solve the equation again, using factorisation, to see if the formula works for this equation.

c. Trying to derive this formula to prove that it always works, Bjorn got stuck along the way. His
attempt his shown below:

ax 2 + bx + c

=

a+^ + 4

X X A

=

Divided by x 2 where i/O

X 1 X

=

Rearranged

-4+ A + 2

x z ex c

=

Divided by c where c/0

x z ex

=

a
c

Subtracted - from both sides

c

x z ex

+

Got stuck

(2.30)

Complete his derivation.

43

Solutions to Exercises in Chapter 2

Solution to Exercise 2.1 (p. 33)

Step 1. This equation has no common factors.

Step 2. The equation is in the required form, with a = 2, b = —5 and c = —12.

Step 3. 2a; 2 — 5a; — 12 has factors of the form:

(2x + s) (x + v)
with s and v constants to be determined. This multiplies out to

2x 2 + (s + 2v)x + sv

(2.31)

(2.32)

We see that sv = —12 and s + 2v = —5. This is a set of simultaneous equations in s and v, but it is
easy to solve numerically. All the options for s and v are considered below.

s

V

s + 2v

2

-6

-10

-2

6

10

3

-4

-5

-3

4

5

4

-3

-2

-4

3

2

6

-2

2

-6

2

-2

Table 2.2

We see that the combination s = 3 and v = —4 gives s + 2v = —5.
Step 4.

(2a; + 3) (a; - 4) = (2.33)

Step 5. If two brackets are multiplied together and give 0, then one of the brackets must be 0, therefore

Therefore, x = — |ora; = 4
Step 6. The solutions to 2a; 2 — 5a; — 12 = are x

2a; + 3 =

x-4 =

| or x = 4.

(2.34)
(2.35)

Solution to Exercise 2.2 (p. 33)

Step 1. Remove the brackets and move all terms to one side.

Step 2.

a z - 3a - 10 =

(a+2)(o-5) =0

(2.36)
(2.37)

44

Step 3.

or

a + 2 =

a-5 =

Solve the two linear equations and check the solutions in the original equation.
Step 4. Therefore, a = — 2 or a = 5

Solution to Exercise 2.3 (p. 33)

Step 1.

36 (6+1) + (6 + 2) (6+1) 4(6+2)

(6 + 2)(6+l) (6 +2) (6 + 1)

Step 2. The denominators are the same, therefore the numerators must be the same.

However, 6^—2 and 6^—1
Step 3.

Step 4.

36 2 + 36 + 6 2 + 36 + 2

= 46+8

46 2 + 26-6

26 2 + 6 - 3

(26 + 3) (6-1) =

26 + 3 = or

6-1 =

b=^ or

6= 1

Step 5. Both solutions are valid

Therefore, 6 = ^ or 6 = 1

Solution to Exercise 2.4 (p. 35)

Step 1.

Step 2.

lOx- 11 =

10a; = 11

(2.38)
(2.39)

(2.40)

(2.41)

(2.42)

Step 3. The coefficient of the x 2 term is 1.

Step 4. The coefficient of the x term is -10. Therefore, half of the coefficient of the x term will be
and the square of it will be (—5) = 25. Therefore:

Step 5.
Step 6.

Step 7.

x - 10x + 25 = 11 + 25

(x - 5) 2 - 36 =

(x - 5) 2 - 36 =

[(x - 5) + 6] [(x - 5) - 6] =

[x + l][x-ll] =

/. x = — 1 or x = 11

(2.43)

(2.44)

(-10) _ r-
2 ~ °

(2.45)

(2.46)

(2.47)

(2.48)

(2.49)

45

Solution to Exercise 2.5 (p. 35)

Step 1.

2a; 2 - 8x - 16 = (2.50)

Step 2.

2x 2 - 8a; = 16 (2.51)

Step 3. The coefficient of the x 2 term is 2. Therefore, divide both sides by 2:

x 2 - 4x = 8 (2.52)

Step 4. The coefficient of the x term is -4; +^ = —2 and (—2) = 4. Therefore:

a; 2 -4a; + 4 = 8 + 4 (2.53)

Step 5.

(x -2) 2 - 12 = (2.54)

Step 6.

[(x - 2) + \/l2~l [(a; - 2) - \/l2~l = (2.55)

Step 7.

[a; - 2 + y/u\ [x - 2 - \/l2~] =

.-. x = 2 - \/l2 or x = 2 + \/l2
Step 8. Leave the left hand side written as a perfect square

Step 4. The two roots of / (a;) = 2a; 2 + 3a; — 7 are x

-b±y/b 2 -4ac

2a

-(3)±V(3) 2 -4(2)(

-7)

2(2)

-3±765

4

-3±\/65

4

- -3+^5 -_ ri -

S-v^

(2.56)

(a;-2) 2 = 12 (2.57)

Step 9. _

x-2 = ±\/l2 (2.58)

Step 10. Therefore x = 2- \/12 or x = 2 + V12
Compare to answer in step 7.

Solution to Exercise 2.6 (p. 37)

Step 1. The expression cannot be factorised. Therefore, the general quadratic formula must be used.
Step 2. From the equation:

a = 2 (2.59)

6 = 3 (2.60)

c=-7 (2.61)

Step 3. Always write down the formula first and then substitute the values of a, b and c.

(2.62)

-b±y f W-

-lac.

2a

-(-5)±V(-5)

2 -4(l)(8)

2(1)

7

46 CHAPTER 2. SOLVING QUADRATIC EQUATIONS

Solution to Exercise 2.7 (p. 37)

Step 1. The expression cannot be factorised. Therefore, the general quadratic formula must be used.
Step 2. From the equation:

a = 1 (2.63)

b = -5 (2.64)

c=8 (2.65)

Step 3.

_ -b±Vb 2 -4ac

2a

(2.66)

Step 4. Since the expression under the square root is negative these are not real solutions (-\/— 7 is not a real
number). Therefore there are no real solutions to the quadratic equation x 2 — 5x + 8 = 0. This means
that the graph of the quadratic function / (x) = x 2 — 5x + 8 has no x-intercepts, but that the entire
graph lies above the x-axis.

Solution to Exercise 2.8 (p. 39)

Step 1. The step before giving the solutions would be:

(x - 13) (x + 5) = (2.67)

Notice that the signs in the brackets are opposite of the given roots.
Step 2.

x 2 - 8x - 65 = (2.68)

Of course, there would be other possibilities as well when each term on each side of the equal to sign
is multiplied by a constant.

Solution to Exercise 2.9 (p. 39)

Step 1. Notice that if x = -§ then 2x + 3 =
Therefore the two brackets will be:

(2x + 3) (x - 4) = (2.69)

Step 2. The equation is:

2a; 2 - 5x - 12 = (2.70)

Chapter 3

3.1 Introduction

Now that you know how to solve quadratic equations, you are ready to learn how to solve quadratic inequal-
ities.

A quadratic inequality is an inequality of the form

ax 2 + bx + c >

ax 2 + bx + c > ,

(3-1)
ax + bx + c <

ax 2 + bx + c <

Solving a quadratic inequality corresponds to working out in what region the graph of a quadratic function
lies above or below the x-axis.

Exercise 3.1: Quadratic Inequality (Solution on p. 49.)

Solve the inequality Ax 2 — Ax + 1 < and interpret the solution graphically.

Exercise 3.2: Solving Quadratic Inequalities (Solution on p. 49.)

Find all the solutions to the inequality x 2 — 5x + 6 > 0.

Exercise 3.3: Solving Quadratic Inequalities (Solution on p. 50.)

Solve the quadratic inequality — x 2 — 3x + 5 > 0.

When working with an inequality where the variable is in the denominator, a different approach is needed.

Exercise 3.4: Non-linear inequality with the variable in the denominator (Solution on
p. 51.)

Solve ^3 < ^

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47

48 CHAPTER 3. SOLVING QUADRATIC INEQUALITIES

Figure 3.1

3.3 End of Chapter Exercises

Solve the following inequalities and show your answer on a number line.

1. Solve: x 2 - x < 12.

2. Solve: 3x 2 > - x + 4

3. Solve: y 2 < -y-2

4. Solve: -t 2 + 2t> - 3

5. Solve: s 2 - 4s > - 6

6. Solve: > 7x 2 - x + 8

7. Solve: > -Ax 2 - x

8. Solve: > 6x 2

9. Solve: 2x 2 + x + 6 <

-3

< 2 and x ^ 3.

-Ai<i.

x — o —

10. Solve for x if

11. Solve for x if

12. Solve for x if

13. Solve for x: ^ff > 3

14. Solve for x: (a _ 3 7 ( 3 a+1) <

15. Solve: (2x - 3) 2 < 4

16. Solve: 2x < ±^E

— X

17. Solve for x: f^f <

18. Solve: x - 2 > ^

— X

19. Solve for x: ^tll^ <

20. Determine all real solutions: %—^- > 1

3— x —

49

Solutions to Exercises in Chapter 3

Solution to Exercise 3.1 (p. 47)

Step 1. Let / (x) = Ax 2 — 4x + 1. Factorising this quadratic function gives / (x) = (2x — 1) .
Step 2.

(2a;-l) 2 <0 (3.2)

Step 3. / (x) = only when x =\.

Step 4. This means that the graph of / (x) = Ax 2 — 4x+ 1 touches the x-axis at x = |, but there are no regions
where the graph is below the x-axis.

Step 5.

Image not finished

Figure 3.2

Solution to Exercise 3.2 (p. 47)

Step 1. The factors of x 2 — 5x + 6 are (x — 3) (x — 2).
Step 2.

x 2 -5x + 6 >
0-3)(x-2) >

(3.3)

Step 3. We need to figure out which values of x satisfy the inequality. From the answers we have five regions
to consider.

Image not finished

Figure 3.3

Step 4. Let / (x) = x 2 — 5x + 6. For each region, choose any point in the region and evaluate the function.

/(*)

sign of / (x)

Region A

x < 2

/(I) =2

+

Region B

x = 2

/(2) =

+

Region C

2 < x < 3

/(2,5) = -2,5

-

Region D

x = 3

/(3) =

+

Region E

x > 3

/(4) = 2

+

Table 3.1

We see that the function is positive for x < 2 and x > 3.

50 CHAPTER 3. SOLVING QUADRATIC INEQUALITIES

Step 5. We see that x 2 — 5x + 6 > is true for x < 2 and x > 3.

Image not finished

Figure 3.4

Solution to Exercise 3.3 (p. 47)

Step 1. Let / (x) = —x 2 — 3x + 5. / (x) cannot be factorised so, use the quadratic formula to determine the
roots of / (x). The x-intercepts are solutions to the quadratic equation

-x 2 - 3x + 5 =

x 2 + 3x - 5 =

-3±V(3) 2 -4(l)(-5)

2 (!) (3.4)

2

3-^29

Xl = 2

to - -3+V29

X2 — 2

Step 2. We need to figure out which values of x satisfy the inequality. From the answers we have five regions
to consider.

Image not finished

Figure 3.5

Step 3. We can use another method to determine the sign of the function over different regions, by drawing
a rough sketch of the graph of the function. We know that the roots of the function correspond to
the x-intercepts of the graph. Let g (x) = —x 2 — 3x + 5. We can see that this is a parabola with a
maximum turning point that intersects the a;-axis at x\ and x^.

Image not finished

Figure 3.6

It is clear that g (x) > for X\ < x < x 2
Step 4. —x 2 — 3a; + 5 > for x\ < x < x 2

51

Image not finished

Figure 3.7

Solution to Exercise 3.4 (p. 47)

Step 1.

Step 2.

x + 3 x — 3

2Qr-3)-QK+3) < n
(x+3)(x-3) - U

1 / x

< (3.5)

x-9
(x+3)(x-3)

<

Image not finished

(3.6)

Step 3.

Figure 3.8

We see that the expression is negative for x < — 3 or 3 < x < 9.
Step 4.

x < - 3 or 3<x<9 (3.7)

52 CHAPTER 3. SOLVING QUADRATIC INEQUALITIES

Chapter 4

Solving simultaneous equations

4.1 Graphical solution 1

4.1.1 Introduction

In grade 10, you learnt how to solve sets of simultaneous equations where both equations were linear (i.e.
had the highest power equal to 1). In this chapter, you will learn how to solve sets of simultaneous equations
where one is linear and one is quadratic. As in Grade 10, the solution will be found both algebraically and
graphically.

The only difference between a system of linear simultaneous equations and a system of simultaneous
equations with one linear and one quadratic equation, is that the second system will have at most two
solutions.

An example of a system of simultaneous equations with one linear equation and one quadratic equation
is:

y - 2x = -4 ,

x z + y = 4

4.1.2 Graphical Solution

The method of graphically finding the solution to one linear and one quadratic equation is identical to
systems of linear simultaneous equations.

4.1.2.1 Method: Graphical solution to a system of simultaneous equations with one linear and

1. Make y the subject of each equation.

2. Draw the graphs of each equation as defined above.

3. The solution of the set of simultaneous equations is given by the intersection points of the two graphs.

For this example, making y the subject of each equation, gives:

y = 2x - 4 ,

A 2 (4 - 2)

y = 4 - X

1 This content is available online at <http://siyavula.cnx.Org/content/m38854/l.l/>.

53

54 CHAPTER 4. SOLVING SIMULTANEOUS EQUATIONS

Plotting the graph of each equation, gives a straight line for the first equation and a parabola for the second
equation.

Image not finished

Figure 4.1

The parabola and the straight line intersect at two points: (2,0) and (-4,-12). Therefore, the solutions to
the system of equations in (4.1) is x = 2, y = and x = —4, y = 12

Exercise 4.1: Simultaneous Equations (Solution on p. 57.)

Solve graphically:

(4.3)
y + 3 X - 9 =

4.1.2.1.1 Graphical Solution

Solve the following systems of equations graphically. Leave your answer in surd form, where appropriate.

1. b 2 -l-a = 0,a + b-5 =

2. x + y-W = 0,x 2 -2-y =

3. 6 - Ax - y = 0, 12 - 2x 2 - y =

4. x + 2y - 14 = 0, x 2 + 2 - y =

5. 2x + 1 - y = 0, 25 - 3a; - x 2 - y =

4.2 Algebraic solution 2
4,2,1 Algebraic Solution

The algebraic method of solving simultaneous equations is by substitution.
For example the solution of

y - 2x = -4 ,

(4.4)
x 2 + y = 4

2 This content is available online at <http://siyavula.cnx.Org/content/m38856/l.l/>.

55

y = 2x — 4 intosecondequation
x 2 + (2x - 4) = 4

x 2 + 2x-8 = (4.5)

Factorisetoget : (x + 4) (x — 2) =

/. the2solutionsfor:rare : x = — 4andx = 2
The corresponding solutions for y are obtained by substitution of the a;- values into the first equation

y = 2(-4)-4 = -12 for a; = -4

V ; (4.6)

and : y = 2 (2) - 4 = for x = 2

As expected, these solutions are identical to those obtained by the graphical solution.

Exercise 4.2: Simultaneous Equations (Solution on p. 57.)

Solve algebraically:

(4.7)
y + 3x - 9 =

4.2.1.1 Algebraic Solution

Solve the following systems of equations algebraically. Leave your answer in surd form, where appropriate.

56

CHAPTER 4. SOLVING SIMULTANEOUS EQUATIONS

1. a + b= 5

a-6 2 + 36-5 =

2. a- 6+1 =

a-6 2 + 56-6 =

3. a (26 + 2) -

a - 26 2 + 36 + 5 =

4. a + 26 -4 =

a- 26 2 - 56+3 =

5. a- 2 + 36=0

a - 9 + 6 2 =

6. a -6-5 =

a - 6 2 =

7. a -6-4 =

a + 26 2 - 12 =

8. a + 6-9 =

a+6 2 - 18 =

9. a -36 + 5 =

a + 6 2 - 46 =

10. a+6-5 =

a - 6 2 + 1 =

11. a- 26- 3 =

a - 36 2 + 4 =

12. a -26=

a- 6 2 - 26 + 3 =

13.a-36 =

a - 6 2 + 4 =

14. a -26- 10 =

a - 6 2 - 56 =

15. a- 36- 1 =

a- 26 2 -6 + 3 =

16. a -36+ 1 =

a - 6 2 =

17. a+ 66-5 =

a - 6 2 - 8 =

18.a-26+l =

a- 26 2 - 126+4 =

19. 2a + 6 - 2 =

8a + 6 2 - 8 =

20. a + 46- 19 =

8a + 56 2 - 101 =

21. a + 46- 18 =

2a + 56 2 -57=

Table 4.1

57

Solutions to Exercises in Chapter 4

Solution to Exercise 4.1 (p. 54)

Step 1. For the first equation:

y - x 2 + 9 =

y = x 2 - 9
and for the second equation:

y + 3x - 9 =

y = -3a; + 9

Image not finished

Step 2.

Figure 4.2

Step 3. The graphs intersect at (—6,27) and at (3,0).

Step 4. The first solution is x = —6 and y = 27. The second solution is x = 3 and y = 0.

Solution to Exercise 4.2 (p. 55)

Step 1.

y + 3x - 9 =

y = -3a; + 9

Step 2.

(-3x + 9)-x 2 + 9 =
x 2 + 3x - 18 =
Factorise to get : (x + 6) (a; — 3) =
.". the 2 solutions or x are : x = —6 and x = 3

Step 3.

y = -3(-6) + 9 = 27 for a; = -6
and : y = —3 (3) + 9 = for x = 3
Step 4. The first solution is x = —6 and y = 27. The second solution is x = 3 and y = 0.

(4i

(4.9)

(4.10)

(4.11)

(4.12)

58 CHAPTER 4. SOLVING SIMULTANEOUS EQUATIONS

Chapter 5

Mathematical Models 1

5.1 Mathematical Models - Grade 11

Up until now, you have only learnt how to solve equations and inequalities, but there has not been much
application of what you have learnt. This chapter builds on this knowledge and introduces you to the idea
of a mathematical model which uses mathematical concepts to solve real-world problems.

Definition 5.1: Mathematical Model

A mathematical model is a method of using the mathematical language to describe the behaviour of
a physical system. Mathematical models are used particularly in the natural sciences and engineer-
ing disciplines (such as physics, biology, and electrical engineering) but also in the social sciences
(such as economics, sociology and political science); physicists, engineers, computer scientists, and
economists use mathematical models most extensively.

A mathematical model is an equation (or a set of equations for the more difficult problems) that describes
a particular situation. For example, if Anna receives R3 for each time she helps her mother wash the dishes
and R5 for each time she helps her father cut the grass, how much money will Anna earn if she helps her
mother 5 times to wash the dishes and helps her father 2 times to wash the car. The first step to modelling
is to write the equation, that describes the situation. To calculate how much Anna will earn we see that she
will earn :

5 x R3 for washing the dishes

+ 2 x R5 for cutting the grass

(5.1)
= R15 +R10

= R25

If however, we ask: "What is the equation if Anna helps her mother x times and her father y times?", then
we have:

Total earned = x x R3 + y x R5 (5.2)

5.2 Real- World Applications: Mathematical Models

Some examples of where mathematical models are used in the real-world are:

lr This content is available online at <http://siyavula.cnx.Org/content/m32649/l.3/>.

59

60 CHAPTER 5. MATHEMATICAL MODELS

1. To model population growth

2. To model effects of air pollution

3. To model effects of global warming

4. In computer games

5. In the sciences (e.g. physics, chemistry, biology) to understand how the natural world works

6. In simulators that are used to train people in certain jobs, like pilots, doctors and soldiers

7. In medicine to track the progress of a disease

5.2,1 Investigation : Simple Models

In order to get used to the idea of mathematical models, try the following simple models. Write an equation
that describes the following real-world situations, mathematically:

1. Jack and Jill both have colds. Jack sneezes twice for each sneeze of Jill's. If Jill sneezes x times, write
an equation describing how many times they both sneezed?

2. It rains half as much in July as it does in December. If it rains y mm in July, write an expression
relating the rainfall in July and December.

3. Zane can paint a room in 4 hours. Billy can paint a room in 2 hours. How long will it take both of
them to paint a room together?

4. 25 years ago, Arthur was 5 more than | as old as Lee was. Today, Lee is 26 less than twice Arthur's
age. How old is Lee?

5. Kevin has played a few games of ten-pin bowling. In the third game, Kevin scored 80 more than in
the second game. In the first game Kevin scored 110 less than the third game. His total score for the
first two games was 208. If he wants an average score of 146, what must he score on the fourth game?

6. Erica has decided to treat her friends to coffee at the Corner Coffee House. Erica paid R54,00 for four
cups of cappuccino and three cups of filter coffee. If a cup of cappuccino costs R3,00 more than a cup
of filter coffee, calculate how much each type of coffee costs?

7. The product of two integers is 95. Find the integers if their total is 24.

Exercise 5.1: Mathematical Modelling of Falling Objects (Solution on p. 65.)

When an object is dropped or thrown downward, the distance, d, that it falls in time, t is described
by the following equation:

s = 5t 2 + v t (5.3)

In this equation, vo is the initial velocity, in m-s -1 . Distance is measured in meters and time is
measured in seconds. Use the equation to find how far an object will fall in 2 s if it is thrown
downward at an initial velocity of 10 m-s -1 .

Exercise 5.2: Another Mathematical Modelling of Falling Objects (Solution on p. 65.)

When an object is dropped or thrown downward, the distance, d, that it falls in time, t is described
by the following equation:

s = 5t 2 + v t (5.4)

In this equation, vo is the initial velocity, in m-s -1 . Distance is measured in meters and time is
measured in seconds. Use the equation find how long it takes for the object to reach the ground if
it is dropped from a height of 2000 m. The initial velocity is m-s -1 .

61

5.2.2 Investigation : Mathematical Modelling

The graph below shows how the distance travelled by a car depends on time. Use the graph to answer the
following questions.

Image not finished

Figure 5.1

1. How far does the car travel in 20 s?

2. How long does it take the car to travel 300 m?

Exercise 5.3: More Mathematical Modelling (Solution on p. 65.)

A researcher is investigating the number of trees in a forest over a period of n years. After
investigating numerous data, the following data model emerged:

Year

Number of trees in hundreds

1

1

2

3

3

9

4

27

Table 5.1

1. How many trees, in hundreds, are there in the SIXTH year if this pattern is continued?

2. Determine an algebraic expression that describes the number of trees in the n th year in the
forest.

3. Do you think this model, which determines the number of trees in the forest, will continue

Exercise 5.4: Setting up an equation (Solution on p. 66.)

Currently the subscription to a gym for a single member is Rl 000 annually while family member-
ship is Rl 500. The gym is considering raising all memberships fees by the same amount. If this
is done then the single membership will cost | of the family membership. Determine the proposed

5.3 End of Chapter Exercises

1. When an object is dropped or thrown downward, the distance, d, that it falls in time, t, is described
by the following equation:

s = 5t 2 + v t (5.5)

In this equation, vo is the initial velocity, in m-s -1 . Distance is measured in meters and time is
measured in seconds. Use the equation to find how long it takes a tennis ball to reach the ground if it
is thrown downward from a hot-air balloon that is 500 m high. The tennis ball is thrown at an initial
velocity of 5 m-s -1 .

62

CHAPTER 5. MATHEMATICAL MODELS

2. The table below lists the times that Sheila takes to walk the given distances.

Time (minutes)

5

10

15

20

25

30

Distance (km)

1

2

3

4

5

6

Table 5.2

Plot the points. If the relationship between the distances and times is linear, find the equation of the
straight line, using any two points. Then use the equation to answer the following questions:

a. How long will it take Sheila to walk 21 km?

b. How far will Sheila walk in 7 minutes?

If Sheila were to walk half as fast as she is currently walking, what would the graph of her distances
and times look like?

3. The power P (in watts) supplied to a circuit by a 12 volt battery is given by the formula P = 127— 0, 5/ 2
where / is the current in amperes.

a. Since both power and current must be greater than 0, find the limits of the current that can be
drawn by the circuit.

b. Draw a graph of P = 12/ — 0, hi 2 and use your answer to the first question, to define the extent
of the graph.

c. What is the maximum current that can be drawn?

d. From your graph, read off how much power is supplied to the circuit when the current is 10

e. At what value of current will the power supplied be a maximum?

4. You are in the lobby of a business building waiting for the lift. You are late for a meeting and wonder
if it will be quicker to take the stairs. There is a fascinating relationship between the number of floors
in the building, the number of people in the lift and how often it will stop: If N people get into a lift
at the lobby and the number of floors in the building is F, then the lift can be expected to stop

F - 1

A'

(5.6)

times.

a. If the building has 16 floors and there are 9 people who get into the lift, how many times is the
lift expected to stop?

b. How many people would you expect in a lift, if it stopped 12 times and there are 17 floors?

5. A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides 3x,
Ax and 5a;. The length of the block is y. The total surface area of the block is 3 600 cm 2 .

Image not finished

Figure 5.2

Show that

300

(5.7)

63

6. A stone is thrown vertically upwards and its height (in metres) above the ground at time t (in seconds)
is given by:

h (t) = 35 - 5t 2 + 30£ (5.8)

Find its initial height above the ground.

7. After doing some research, a transport company has determined that the rate at which petrol is
consumed by one of its large carriers, travelling at an average speed of x km per hour, is given by:

P (x) = 1 litresperkilometre (5-9)

Assume that the petrol costs R4,00 per litre and the driver earns R18,00 per hour (travelling time).
Now deduce that the total cost, C, in Rands, for a 2 000 km trip is given by:

C{x)=™™ + 40x (5.10)

x

8. During an experiment the temperature T (in degrees Celsius), varies with time t (in hours), according
to the formula:

T(t) = 30 + 4t- -t 2 te[l;10] (5.11)

a. Determine an expression for the rate of change of temperature with time.

b. During which time interval was the temperature dropping?

9. In order to reduce the temperature in a room from 28°C, a cooling system is allowed to operate for 10
minutes. The room temperature, T after t minutes is given in °C by the formula:

T = 28 - 0, 008t 3 - 0, 16* wheret e [0; 10] (5.12)

a. At what rate (rounded off to TWO decimal places) is the temperature falling when t = 4 minutes?

b. Find the lowest room temperature reached during the 10 minutes for which the cooling system
operates, by drawing a graph.

10. A washing powder box has the shape of a rectangular prism as shown in the diagram below. The box
has a volume of 480 cm 3 , a breadth of 4 cm and a length of x cm.

Image not finished

Figure 5.3

Show that the total surface area of the box (in cm 2 ) is given by:

A= 82; + 960x~ 1 + 240 (5.13)

64 CHAPTER 5. MATHEMATICAL MODELS

5.3.1 Simulations

A simulation is an attempt to model a real-life situation on a computer so that it can be studied to see
how the system works. By changing variables, predictions may be made about the behaviour of the system.
Simulation is used in many contexts, including the modeling of natural systems or human systems in order
to gain insight into their functioning. Other contexts include simulation of technology for performance
optimization, safety engineering, testing, training and education. Simulation can be used to show the eventual
real effects of alternative conditions and courses of action.

Simulation in education Simulations in education are somewhat like training simulations. They focus
on specific tasks. In the past, video has been used for teachers and education students to observe, problem
solve and role play; however, a more recent use of simulations in education is that of animated narrative
vignettes ( ANV) . ANVs are cartoon- like video narratives of hypothetical and reality-based stories involving
classroom teaching and learning. ANVs have been used to assess knowledge, problem solving skills and
dispositions of children and pre-service and in-service teachers.

65

Solutions to Exercises in Chapter 5

Solution to Exercise 5.1 (p. 60)

Step 1. We are given an expression to calculate distance traveled by a falling object in terms of initial velocity
and time. We are also given the initial velocity and time and are required to calculate the distance
traveled.

Step 2.

• vq = 10 m ■ s

• t = 2s

• s =? m

Step 3.

s = 5£ 2 + v t

= 5(2) 2 + (10)(2)

5 (4) + 20 (5.14)

20 + 20
40

Step 4. The object will fall 40 m in 2 s if it is thrown downward at an initial velocity of 10 m-s -1 .

Solution to Exercise 5.2 (p. 60)

Step 1. We are given an expression to calculate distance travelled by a falling object in terms of initial velocity
and time. We are also given the initial velocity and distance travelled and are required to calculate
the time it takes the object to travel the distance.

Step 2.

• Vq = m ■ s

• t=ls

• s = 2000 m

Step 3.

s = 5t + v t
2000 = 5t 2 + (0) (2)
2000 = 5£ 2

(5.15)

2 2000 v '

~ 5

400

.-. t = 20 s

Step 4. The object will take 20 s to reach the ground if it is dropped from a height of 2000 m.

Solution to Exercise 5.3 (p. 61)

Step 1. The pattern is 3°; 3 1 ; 3 2 ; 3 3 ; ...

Therefore, three to the power one less than the year.
Step 2.

year 6 : 3 hundreds = 243 hundreds = 24300 (5.16)

Step 3.

number of trees = 3™ _ hundreds (5-17)

Step 4. No

The number of trees will increase for some time. Yet, eventually the number of trees will not increase
any more. It will be limited by factors such as the limited amount of water and nutrients available in
the ecosystem.

66

Solution to Exercise 5.4 (p. 61)

Step 1. Let the proposed increase be x.

CHAPTER 5. MATHEMATICAL MODELS

Step 2.
Step 3.

Now

After increase

Single

1 000

1 000+x

Family

1 500

1 500+x

Table 5.3

1 000 + x= -(1 500 + x)

7

7 000 + 7x = 7 500 + 5x
2x = 500

x = 250

(5.18)

(5.19)

Step 4. Therefore the increase is R250.

Chapter 6

6.1 Introduction

In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about

6.2 Functions of the Form y = a(x + p) + q

This form of the quadratic function is slightly more complex than the form studied in Grade 10, y = ax 2 + q.
The general shape and position of the graph of the function of the form / (x) = a(x + p) + q is shown in
Figure 6.1.

Image not finished

Figure 6.1: Graph of / (a;) = \{x + 2) 2 - 1

6.2.1 Investigation : Functions of the Form y = a(x + p) + q

1.

On

the same set of axes,

plot

the following

graphs:

a.

a (x) =

:(x-2f

b.

b(x) =

(x-lf

c.

c (x) =

x 2

d.

d{x) =

■{x+lf

c.

e{x) =

{x + 2f

Use

your results to deduce the effect of p.

2.

On

the same set of axes,

plot

the following

graphs:

a.

/(*) =

--{x-2) 2 + l

b.

g{x) =

-{x-lf + 1

lr This content is available online at <http://siyavula.cnx.Org/content/m30843/l.4/>.

67

68

CHAPTER 6. QUADRATIC FUNCTIONS AND GRAPHS

c. h(x) = x + 1

d. j(x) = {x+lf + l

e. k{x) = (x + 2) 2 + 1

Use your results to deduce the effect of q.
3. Following the general method of the above activities, choose your own values of p and q to plot 5
different graphs (on the same set of axes) of y = a(x + p) + q to deduce the effect of a.

From your graphs, you should have found that o affects whether the graph makes a smile or a frown. If
a < 0, the graph makes a frown and if a > then the graph makes a smile. This was shown in Grade 10.

You should have also found that the value of q affects whether the turning point of the graph is above
the x-axis (q < 0) or below the x-axis (q > 0).

You should have also found that the value of p affects whether the turning point is to the left of the
y-axis (p > 0) or to the right of the y-axis (p < 0).

These different properties are summarised in Table 6.1. The axes of symmetry for each graph is shown
as a dashed line.

p <

a >

i

V

,, i

Figure 6.2

continued on next page

69

q < o

1 i

'

1

- 1

Figure 6.6

Table 6.1: Table summarising general shapes and positions of functions of the form y = a(x + p) + q. The

axes of symmetry are shown as dashed lines.

Phet simulation for graphing

< equat ion-grapher . s wf >

Figure 6.10

6.2,2 Domain and Range

For / (x) = a(x + p) + q, the domain is {x : x G K} because there is no value of x G K for which / (x) is
undefined.

The range of / (x) = a(x + p) + q depends on whether the value for a is positive or negative. We will
consider these two cases separately.

If a > then we have:

(x + p) > (The square of an expression is always positive) a(x + p) > (6.1)

(Multiplication by a positive number maintains the nature of the inequality) a(x + p) +
q > qf(x) > q

This tells us that for all values of x, f (x) is always greater than or equal to q. Therefore if a > 0, the
range of / (x) = a(x + p) + q is {/ (x) : f (x) G [q, oo)}.

Similarly, it can be shown that if a < that the range of / (x) = a(x + p) +q is {/ (x) : f (x) € (— oo, q}}.
This is left as an exercise.

For example, the domain of g (x) = (x — 1) + 2 is {x : x G M} because there is no value of x G M for
which g (x) is undefined. The range of g (x) can be calculated as follows:

(x-p) >

{x + pf + 2 > 2

g{x) > 2
Therefore the range is {g (x) : g (x) G [2, oo)}.

(6.2)

70 CHAPTER 6. QUADRATIC FUNCTIONS AND GRAPHS

6.2.2.1 Domain and Range

1. Given the function / (x) = (x — 4) — 1. Give the range of / (x).

2. What is the domain of the equation y = 2x 2 — 5a; — 18 ?

6.2,3 Intercepts

,2

For functions of the form, y = a(x + p) + q, the details of calculating the intercepts with the x and y axes
is given.

The y-intercept is calculated as follows:

y = a (x + p) 2 + q
y tn t = a{0 + pf + q (6.3)

= ap 2 + q

If p = 0, then y int = q.

2

For example, the y-intercept of g (x) = (x — 1) + 2 is given by setting x = to get:

g(x) = (x-lf + 2

Vint = (0-1

,2

(-1) +2 (6.4)

1 + 2
3

The x-intercepts are calculated as follows:

y = a(x + pf + q

= a(x int +p) + q

a{x int +pf = -q (6-5)

Xint+P = ii/-"?

Xint — ^ 1

However, (6.5) is only valid if — - > which means that either q < or a < but not both. This is
consistent with what we expect, since if q > and a > then —-is negative and in this case the graph
lies above the x-axis and therefore does not intersect the x-axis. If however, q > and a < 0, then — - is
positive and the graph is hat shaped with turning point above the x-axis and should have two x-intercepts.
Similarly, if q < and a > then — ^ is also positive, and the graph should intersect with the x-axis twice.
For example, the x-intercepts of g (x) = (x — 1) +2 are given by setting y = to get:

g{x) = {x-lf + 2

,2

= (Xint-l) '+2 (6.6)

"^ — \Xint ^

,2

2

which has no real solutions. Therefore, the graph of g (x) = (x — 1) +2 does not have any x-intercepts.

71

6.2.3.1 Intercepts

1. Find the x- and y-intercepts of the function / (x) = (x — 4) — 1.

2. Find the intercepts with both axes of the graph of / (x) = x 2 — 6x + 8.

3. Given: / (x) = —x 2 + Ax — 3. Calculate the x- and y-intercepts of the graph of /.

6.2.4 Turning Points

The turning point of the function of the form / (x) = a(x + p) + q is given by examining the range of the
function. We know that if a > then the range of / (x) = a(x + p) + q is {/ (x) : f (x) G [q, oo)} and if
a < then the range of / (x) = a(x + p) + q is {/ (aj) : / (x) G (— oo, q]}.

So, if a > 0, then the lowest value that / (x) can take on is q. Solving for the value of x at which / (x) = q
gives:

q =

= a(x + p) + q

o =

a(x + p)

o =

(x + pf

o =

x + p

(6.7)

x = —p

:.x = — p at / (x) = q. The co-ordinates of the (minimal) turning point is therefore (— p, q).

Similarly, if a < 0, then the highest value that / (x) can take on is q and the co-ordinates of the (maximal)
turning point is (—p,q).

6.2.4.1 Turning Points

1. Determine the turning point of the graph of / (x) = x 2 — 6x + 8 .

2. Given: / (x) = —x 2 + Ax — 3. Calculate the co-ordinates of the turning point of /.

3. Find the turning point of the following function by completing the square: y = i(x + 2) — 1.

6.2.5 Axes of Symmetry

There is only one axis of symmetry for the function of the form / (x) = a(x + p) + q. This axis passes
through the turning point and is parallel to the y-axis. Since the x-coordinate of the turning point is x = —p,
this is the axis of symmetry.

6.2.5.1 Axes of Symmetry

1. Find the equation of the axis of symmetry of the graph y = 2x 2 — 5x — 18.

2. Write down the equation of the axis of symmetry of the graph of y = 3(x — 2) +1.

3. Write down an example of an equation of a parabola where the y-axis is the axis of symmetry.

6.2.6 Sketching Graphs of the Form / (x) = a(x + p) + q

In order to sketch graphs of the form / (x) = a(x + p) + q, we need to determine five characteristics:

1. sign of a

2. domain and range

3. turning point

72 CHAPTER 6. QUADRATIC FUNCTIONS AND GRAPHS

4. y-intercept

5. a;- intercept

For example, sketch the graph of g (x) = — \(x + 1) — 3. Mark the intercepts, turning point and axis of
symmetry.

Firstly, we determine that a < 0. This means that the graph will have a maximal turning point.

The domain of the graph is {x : x e K} because / (x) is defined for all i£i The range of the graph is
determined as follows:

(61

Therefore the range of the graph is {/ (x) : f (x) € (—00, —3]}.

Using the fact that the maximum value that / (x) achieves is -3, then the y-coordinate of the turning
point is -3. The ^-coordinate is determined as follows:

(x+1) 2

>

-\(x+l) 2

<

i(a;+l) 2 -3

<

-3

.-./(*)

<

-3

±(:r+l) 2 -3 = -3
i(x+l) 2 -3 + 3 =
Ux+lf =

2

2

Divide both sides by — \ : (x + 1) =
Take square root of both sides : x + 1 =

(6.9)

-1

The coordinates of the turning point are: ( — 1, —3).

The y-intercept is obtained by setting x = 0. This gives:

Vint

=

-i(0+l) 2 -3

=

-1(1) -3

=

-1 -3

2 a

=

-i - 3
2 6

=

7
2

i = 0.

This

gives:

= -

~2\ X int + 1) — 3

3

=

2 y^int ' ^ )

3-2

=

(Xint + 1)

-6

=

(Xint + 1)

(6.10)

(6.11)

which has no real solutions. Therefore, there are no x-intercepts.

We also know that the axis of symmetry is parallel to the y-axis and passes through the turning point.

73

Image not finished

Figure 6.11: Graph of the function f (x) — — \{x + 1) — 3

Figure 6.12

6.2.6.1 Sketching the Parabola

1. Draw the graph of y = 3(x — 2) +1 showing all the intercepts with the axes as well as the coordinates
of the turning point.

2. Draw a neat sketch graph of the function defined by y = ax 2 + bx + c if a > 0; b < 0; b 2 = Aac.

6.2,7 Writing an equation of a shifted parabola

Given a parabola with equation y = x 2 — 2x — 3. The graph of the parabola is shifted one unit to the right.
Or else the y-axis shifts one unit to the left i.e. x becomes x — 1. Therefore the new equation will become:

y = (x- if -2 (x- 1) -3

= x 2 -2x+ 1-2X + 2-3 (6.12)

= x 2 — Ax

If the given parabola is shifted 3 units down i.e. y becomes y + 3. The new equation will be:
(Notice the x-axis then moves 3 units upwards)

y + 3 = a; 2 -2a;-3 ,

(6-13)
y = x — 2x — 6

6.3 End of Chapter Exercises

,2

1. Show that if a < 0, then the range of / (x) = a(x + p) + q is {/ (x) : / (x) € (—00, q]}.

2. If (2,7) is the turning point of / (x) = — 2x 2 — Aax + k, find the values of the constants o and k.

3. The graph in the figure is represented by the equation / (x)
point are (3,9). Show that a = — 1 and 6=6.

74 CHAPTER 6. QUADRATIC FUNCTIONS AND GRAPHS

Image notjtnished

Figure 6.13

4. Given: y = x 2 — 2x + 3. Give the equation of the new graph originating if:

a. The graph of / is moved three units to the left.

b. The x-axis is moved down three units.

5. A parabola with turning point (-1,-4) is shifted vertically by 4 units upwards. What are the coordinates
of the turning point of the shifted parabola?

Chapter 7

Hyperbolic Functions and Graphs 1

7.1 Introduction

In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about
the graphs of functions.

7.2 Functions of the Form y = -7 — \- q

u x+p *

This form of the hyperbolic function is slightly more complex than the form studied in Grade 10.

Image not finished

Figure 7.1: General shape and position of the graph of a function of the form / (x) = — ^ — \- q. The
asymptotes are shown as dashed lines.

7.2.1 Investigation : Functions of the Form y = -7-

° a x+p

1. On the same set of axes, plot the following graphs:

a.

a (x) =

-2

x+l

+ 1

b.

b(x) =

-1

x+l

+ 1

c.

c(x) =

U

x+l

+ 1

d.

d{x) =

1

x+l

+ 1

0.

e{x) =

2
x+l

+ 1

Use

ults

to deduce the effect of a.

2. On

the same set of axes,

plot

the following

graphs:

a.

/(*) =

1
x-2

+ 1

b.

g{x) =

1
x-1

+ 1

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75

76

CHAPTER 7. HYPERBOLIC FUNCTIONS AND GRAPHS

c. h (x)

d. j (x) ■■

1

x+0

1
x+1

e. k(x) = jq^
Use your results to deduce the effect of p.
3. Following the general method of the above activities, choose your own values of a and p to plot 5

different graphs of y

x+p

q to deduce the effect of q.

You should have found that the sign of a affects whether the graph is located in the first and third quadrants,
or the second and fourth quadrants of Cartesian plane.

You should have also found that the value of p affects whether the a;-intercept is negative (p > 0) or
positive (p < 0).

You should have also found that the value of q affects whether the graph lies above the x-axis (q > 0) or
below the x-axis (q < 0).

These different properties are summarised in Table 7.1. The axes of symmetry for each graph is shown
as a dashed line.

p <

a >

Figure 7.2

continued on next page

77

q <

t '

1

1

■*-=-=<

..lk^

■ 'ni

Figure 7.6

Table 7.1: Table summarising general shapes and positions of functions of the form y = ^ — h q. The axes

of symmetry are shown as dashed lines.

7.2,2 Domain and Range

For y = -^rt — I- q, the function is undefined for x = —p. The domain is therefore {i:i£l,i/ — p).
We see that y = ^r — h q can be re-written as:

y

—

— h

x+p

y- q

=

a
x+p

If x±

—p then :

(y-

- q){x + p)

=

a

x + p

=

a

(7.1)

This shows that the function is undefined at y = q. Therefore the range of / (x) = -^— + q is {/ (x) : f (x) s

R,f(x)?q.

For example, the domain of g (x) = -^-y +2is {i:i£R,i/ -1} because g (x) is undefined at x = — 1.

y
(v-2)

( V -2)(a:+l)
(ar+1)
We see that g (x) is undefined at y = 2. Therefore the range is {g (x) : g (x) € (— oo, 2) U (2, oo)}.

k+1 '

2
x+1

2

y-2

(7.2)

7.2.2.1 Domain and Range

1. Determine the range of y = - + 1.

2. Given:/ (x) = -^^ + 4. Write down the domain of /.

3. Determine the domain of y ■■

x+l

7.2,3 Intercepts

For functions of the form, y = -^r — h q, the intercepts with the x and y axis are calculated by setting x =
for the y-intercept and by setting y = for the x-intercept.

78 CHAPTER 7. HYPERBOLIC FUNCTIONS AND GRAPHS

The y-intercept is calculated as follows:

y = ^ + i

Vint = 0+^+9 ( 7 - 3 )

-+q
For example, the y-intercept of g (x) = -^-^ + 2 is given by setting x = to get:

(7.4)

V =

S+T + 2

! int —

-2 i_ 2

0+1 + z

=

f + 2

= 2 + 2
4
The x-intercepts are calculated by setting y = as follows:

y= ittp + i

Xint+P

Xint+P

-q

a

=

-q{x int +p)

■

Xint+P

a

-'!

%int

a

-,-p

2
x+l

+ 2 is g:

tven by

setting

y

a

2 i_2

+i + z

2 4-

Kit

,t + l +

-2

=

2

x»„t + l

2 (X int + 1)

=

2

%int \ J-

=

2
-2

•Eint

=

-1- 1

•^int

=

-2

(7.5)

(7.6)

79

7.2.3.1 Intercepts

1. Givemh (x) = -^-^ — 2. Determine the coordinates of the intercepts of h with the x- and y-axes.

2. Determine the x-intercept of the graph of y = - + 2. Give the reason why there is no y-intercept for
this function.

7.2,4 Asymptotes

There are two asymptotes for functions of the form y = -^ — \- q. They are determined by examining the
domain and range.

We saw that the function was undefined at x = —p and for y = q. Therefore the asymptotes are x = —p
and y = q.

For example, the domain of g (x) = ^|-j- + 2is {i:i£l,i/ - 1} because g (x) is undefined at x = — 1.
We also see that g (x) is undefined at y = 2. Therefore the range is {g (x) : g (x) s (—00, 2) U (2, oo)}.

From this we deduce that the asymptotes are at x = — 1 and y = 2.

7.2.4.1 Asymptotes

1. Given:/i (x) = -A^ — 2. Determine the equations of the asymptotes of h.

2. Write down the equation of the vertical asymptote of the graph y = —^-r.

7.2.5 Sketching Graphs of the Form / (x) = ^- + q

In order to sketch graphs of functions of the form, / (x) = -^— + q, we need to calculate four characteristics:

1. domain and range

2. asymptotes

3. y-intercept

4. a;-intercept

For example, sketch the graph of g (x) = — ^ + 2. Mark the intercepts and asymptotes.

We have determined the domain to be {i : i e l,i / -1} and the range to be {g (x) : g (x) €
(— oo, 2) U (2, oo)}. Therefore the asymptotes are at x = — 1 and y = 2.

The y-intercept is yi nt = 4 and the x-intercept is X{ n t = — 2.

Image notjtnished

Figure 7.10: Graph of g (x) - ^ + 2.

80 CHAPTER 7. HYPERBOLIC FUNCTIONS AND GRAPHS

7.2.5.1 Graphs

1. Draw the graph of y = - + 2. Indicate the horizontal asymptote.

2. Given:/i (x) = -^ — 2. Sketch the graph of h showing clearly the asymptotes and ALL intercepts with
the axes.

3. Draw the graph of y = ^ and y = — ^4r[ + 3 on the same system of axes.

4. Draw the graph of y = x _^ 2 5 + 2. Explain your method.

5. Draw the graph of the function defined by y = -^^ + 4. Indicate the asymptotes and intercepts with
the axes.

7.3 End of Chapter Exercises

1. Plot the graph of the hyperbola defined by y = - for —4 < x < 4. Suppose the hyperbola is shifted 3
units to the right and 1 unit down. What is the new equation then ?

2. Based on the graph of y = -, determine the equation of the graph with asymptotes y = 2 and x = 1
and passing through the point (2; 3).

Chapter 8

Exponential Functions and Graphs 1

8.1 Introduction

In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about
the graphs of exponential functions.

8.2 Functions of the Form y = ab^ x+p ^ + q for b >

This form of the exponential function is slightly more complex than the form studied in Grade 10.

Image not finished

Figure 8.1: General shape and position of the graph of a function of the form / (x) — ab^ x+p ' + q.

8.2.1 Investigation : Functions of the Form y = ab^ x+p ^ + q

1. On the same set of axes, with — 5 < x < 3 and —35 < y < 35, plot the following graphs:

a. f(x) = -2-2< a: + 1 ) + 1

b. g(x) = -1 -2 {x+ V + 1

c. h\x) = 0-2 ( - x+1 '> + 1

d. j (x) = 1 • 2 {x+1 ^ + 1

e. k{x) = 2-2( x+1 ) + 1

Use your results to understand what happens when you change the value of a. You should find that
the value of a affects whether the graph curves upwards (a > 0) or curves downwards (a < 0). You
should also find that a larger value of o (when a is positive) stretches the graph upwards. However,
when a is negative, a lower value of a (such as -2 instead of -1) stretches the graph downwards. Finally,
note that when a = the graph is simply a horizontal line. This is why we set a / in the original
definition of these functions.

2. On the same set of axes, with —3 < x < 3 and — 5 < y < 20, plot the following graphs:

lr This content is available online at <http://siyavula.cnx.Org/content/m30856/l.3/>.

81

82

CHAPTER 8. EXPONENTIAL FUNCTIONS AND GRAPHS

a.

f(x) = l

. 2 (s+i) _ 2

b.

g(x) = l

2 (x+i) _ 1

c.

h(x) = l

2(^+1) + o

d.

j (x) = 1

2 (x+i) + x

0.

k(x) = l

2 (x+i) + 2

Use your results to understand what happens when you change the value of q. You should find that
when q is increased, the whole graph is translated (moved) upwards. When q is decreased (poosibly
even made negative), the graph is translated downwards.
3. On the same set of axes, with — 5 < x < 3 and —35 < y < 35, plot the following graphs:

f{x) = -2 ■ 2 ( - x+1 '> + 1
g(x) = -1 -2 {x+ V + 1
h\x) = 0-2( x+1 ) + 1

a.
b.
c.

d.

e.

2 (x+i) + 1

3 0) =

k{x) = 2-2< 2: + 1 ) + l

Use your results to understand what happens when you change the value of a. You should find that
the value of a affects whether the graph curves upwards (a > 0) or curves downwards (a < 0). You
should also find that a larger value of o (when a is positive) stretches the graph upwards. However,
when a is negative, a lower value of a (such as -2 instead of -1) stretches the graph downwards. Finally,
note that when a = the graph is simply a horizontal line. This is why we set a / in the original
definition of these functions.
4. Following the general method of the above activities, choose your own values of a and q to plot 5 graphs
of y = ab( x+p } + q on the same set of axes (choose your own limits for x and y carefully) . Make sure
that you use the same values of a, b and q for each graph, and different values of p. Use your results
to understand the effect of changing the value of p.

These different properties are summarised in Table 8.1.

p <

a >

Figure 8.2

continued on next page

83

Figure 8.6

1.1)

Table 8.1: Table summarising general shapes and positions of functions of the form y = ab^ x+p ^ + q.

8.2.2 Domain and Range

For y = ab( x+p ) + q, the function is defined for all real values of x. Therefore, the domain is {x : x £ R}.
The range of y = ab^ x+p ^ + q is dependent on the sign of a.
If a > then:

b(*+p) > o

a ■ b (x+ P) >

a ■ 6( x +p) + q > q

/0) > q

Therefore, if o > 0, then the range is {/ (x) : f (x) e [q, oo)}. In other words / (a;) can be any real number
greater than q.
If a < then:

b(*+p) > o

a ■ b {x+ P) <

a ■ &( x+ p) + q < q

f(x) < q

1.2)

Therefore, if a < 0, then the range is (—00,(7), meaning that / (x) can be any real number less than q.
Equivalently, one could write that the range is {y G R : y < q}.

For example, the domain of g (x) = 3 • 2 X+1 + 2 is {x : x e R}. For the range,

2 X+1 >

3 • 2 X+1 >

3-2 a; + 1 + 2 > 2

Therefore the range is {g (x) : g (x) € [2, 00)}.

*.3)

84 CHAPTER 8. EXPONENTIAL FUNCTIONS AND GRAPHS

8.2.2.1 Domain and Range

1. Give the domain of y = 3 X .

2. What is the domain and range of / (x) = 2 X ?

3. Determine the domain and range of y = (1,5) .

8.2,3 Intercepts

For functions of the form, y = ab^ x+p ^ + q, the intercepts with the x- and y-axis are calculated by setting
x = for the y-intercept and by setting y = for the ^-intercept.
The y-intercept is calculated as follows:

y = ab^+P^ + q
Vint = ab(°+rt + q (8.4)

= abP + q

For example, the y-intercept of g (x) = 3 ■ 2 X+1 + 2 is given by setting x = to get:

y =

3 • 2 X+1 + 2

Vint —

3 • 2 0+1 + 2

=

3 ■2 1 + 2

=

3-2 + 2

=

8

The x-intercepts are calculated by setting y = as

follows:

y =

ab {x+ rt -

o =

= &(3int+p)

a 5(z 4 „t+p) =

-q

t)(xi nt +p) _

_ _£

(8.5)

3.6)

Since b > (this is a requirement in the original definition) and a positive number raised to any power is
always positive, the last equation above only has a real solution if either a < or q < (but not both).
Additionally, a must not be zero for the division to be valid. If these conditions are not satisfied, the graph
of the function of the form y = ab^ x+p ' + q does not have any x-intercepts.

For example, the x-intercept of g (x) = 3 • 2 X+1 + 2 is given by setting y = to get:

3.7)

3 • 2 Xi " t+1

which has no real solution. Therefore, the graph of g (x) = 3 • 2 X+1 + 2 does not have a x-intercept. You
will notice that calculating g (x) for any value of x will always give a positive number, meaning that y will
never be zero and so the graph will never intersect the x-axis.

y

=

3-2 rE+1 + 2

=

3 . 2 x ^+ l + 2

2

=

3 . 2 z-*+i

-l

=

-2
2

85

8.2.3.1 Intercepts

1. Give the y-intercept of the graph of y = b x + 2.

2. Give the x- and y-intercepts of the graph of y = |(1, 5) x — 0, 75.

8.2.4 Asymptotes

Functions of the form y = aU x+p ^ + q always have exactly one horizontal asymptote.

When examining the range of these functions, we saw that we always have either y < q or y > q for all
input values of x. Therefore the line y = q is an asymptote.

For example, we saw earlier that the range of g (x) = 3 • 2 X+1 + 2 is (2, oo) because g (x) is always greater
than 2. However, the value of g (x) can get extremely close to 2, even though it never reaches it. For example,
if you calculate g (— 20), the value is approximately 2.000006. Using larger negative values of x will make
g (x) even closer to 2: the value of g (—100) is so close to 2 that the calculator is not precise enough to know
the difference, and will (incorrectly) show you that it is equal to exactly 2.

From this we deduce that the line y = 2 is an asymptote.

8.2.4.1 Asymptotes

1. Give the equation of the asymptote of the graph of y = 3 X — 2.

2. What is the equation of the horizontal asymptote of the graph of y = 3(0, 8) x ~ — 3 ?

8.2.5 Sketching Graphs of the Form / (x) = ab^ x+ ^ + q

In order to sketch graphs of functions of the form, / (x) = ab^ x+p ^ + q, we need to determine four character-
istics:

1. domain and range

2. y-intercept

3. ^-intercept

For example, sketch the graph of g (x) = 3 • 2 X+1 + 2. Mark the intercepts.

We have determined the domain to be {x : x e R} and the range to be {g (x) : g (x) s (2, oo)}.
The y-intercept is y int = 8 and there is no x- intercept.

Image not finished

Figure 8.10: Graph of g (x) = 3 • 2 X+1 + 2.

8.2.5.1 Sketching Graphs

1. Draw the graphs of the following on the same set of axes. Label the horizontal asymptotes and y-
intercepts clearly.

a. y = b x + 2

b. y = b x+2

86

CHAPTER 8. EXPONENTIAL FUNCTIONS AND GRAPHS

c. y

d. y

2b x

2b x+2

a. Draw the graph of / (x) = 3 X .

b. Explain where a solution of 3 X = 5 can be read off the graph.

8.3 End of Chapter Exercises

1. The following table of values has columns giving the y- values for the graph y = a x , y
y = a x + 1. Match a graph to a column.

a x+1 and

X

A

B

C

-2

7,25

6,25

2,5

-1

3,5

2,5

1

2

1

0,4

1

1,4

0,4

0,16

2

1,16

0,16

0,064

Table 8.2

2. The graph of / (x) = 1 + a.2 x (a is a constant) passes through the origin.

a. Determine the value of a.

b. Determine the value of / (—15) correct to FIVE decimal places.

c. Determine the value of x, if P (x; 0, 5) lies on the graph of /.

d. If the graph of / is shifted 2 units to the right to give the function h, write down the equation of

h.

3. The graph of / (a;) = a.b x {a =£ 0) has the point P(2;144) on /.

a. If b = 0, 75, calculate the value of a.

b. Hence write down the equation of /.

c. Determine, correct to TWO decimal places, the value of /(13).

d. Describe the transformation of the curve of / to h if h (x) = f (— x).

Chapter 9

9.1 Introduction

In Grade 10, we investigated the idea of average gradient and saw that the gradient of some functions varied
over different intervals. In Grade 11, we further look at the idea of average gradient, and are introduced to
the idea of a gradient of a curve at a point.

We saw that the average gradient between two points on a curve is the gradient of the straight line passing
through the two points.

Image not finished

Figure 9.1: The average gradient between two points on a curve is the gradient of the straight line that
passes through the points.

What happens to the gradient if we fix the position of one point and move the second point closer to the
fixed point?

9.2.1 Investigation : Gradient at a Single Point on a Curve

The curve shown is defined by y = —2x 2 — 5. Point B is fixed at coordinates (0,-5). The position of point
A varies. Complete the table below by calculating the y-coordinates of point A for the given ^-coordinates
and then calculate the average gradient between points A and B.

1 This content is available online at <http://siyavula.cnx.Org/content/m32651/l.3/>.

87

88

CHAPTER 9. GRADIENT AT A POINT

X A

VA

-2

-1.5

-1

-0.5

0.5

1

1.5

2

Table 9.1

Image not finished

Figure 9.2

What happens to the average gradient as A moves towards B? What happens to the average gradient as
A moves away from B? What is the average gradient when A overlaps with B?

In Figure 9.3, the gradient of the straight line that passes through points A and C changes as A moves
closer to C. At the point when A and C overlap, the straight line only passes through one point on the curve.
Such a line is known as a tangent to the curve.

89

Image not finished

(a)

Image not finished

(b)

Image not finished

(c)

Image not finished

(d)

Figure 9.3: The gradient of the straight line between A and C changes as the point A moves along the
curve towards C. There comes a point when A and C overlap (as shown in (c)). At this point the line is
a tangent to the curve.

We therefore introduce the idea of a gradient at a single point on a curve. The gradient at a point on a
curve is simply the gradient of the tangent to the curve at the given point.

Exercise 9.1: Average Gradient (Solution on p. 91.)

Find the average gradient between two points P(a; g (a)) and Q(a + h;g (a + hj) on a curve g (x) =
x 2 . Then find the average gradient between P(2;<7(2)) and Q(4;g(4)). Finally, explain what
happens to the average gradient if P moves closer to Q.

We now see that we can write the equation to calculate average gradient in a slightly different manner. If
we have a curve defined by / (x) then for two points P and Q with P(a; / (a)) and Q(a + h; f (a + h)), then
the average gradient between P and Q on / (x) is:

2/2-2/1 _ f(a+h)-f(a)

X2-X1 (a+h)-(a) (Q \\

_ f(g+h)-f(a) ^- L >

~ h

This result is important for calculating the gradient at a point on a curve and will be explored in greater

9.3 End of Chapter Exercises

1. a. Determine the average gradient of the curve / (x) = x (x + 3) between x = A[U+0096]5 and

x = i[U+0096]3.
b. Hence, state what you can deduce about the function / between x = A [U+0096] 5 and x =
i[U+0096]3.

2. A(l;3) is a point on / (x) = 3a; 2 .

a. Determine the gradient of the curve at point A.

b. Hence, determine the equation of the tangent line at A.

3. Given: / (ar) = 2x 2 .

90 CHAPTER 9. GRADIENT AT A POINT

a. Determine the average gradient of the curve between x = — 2 and x = 1.

b. Determine the gradient of the curve of / where x = 2.

91

Solutions to Exercises in Chapter 9

Solution to Exercise 9.1 (p. 89)

Step 1.

xi = a (9.2)

x 2 = a+h (9.3)

Step 2. Using the function g (x) = x 2 , we can determine:

Vl =g (a) = a 2 (9.4)

2/2 = g(a + h)

Step 3.

=

{a+hf

=

a 2

+ 2ah+ h 2

Vi-V\

=

(«

<. 2 +2a/i+/i 2 )-(a 2 )

x 2 —xi

(o+/i)-(o)

=

a 2 +2aft+/i 2 -a 2
a+/l— a

=

2aft+h 2
ft

=

/i(2a+h)
ft

2a + ft,

(9.5)

(9.6)

The average gradient between P(o; g (a)) and Q(a + ft; 5 (a + ft)) on the curve g (x) = x 2 is 2a + ft.
Step 4. We can use the result in (9.6), but we have to determine what a and ft are. We do this by looking at
the definitions of P and Q. The ^-coordinate of P is a and the x-coordinate of Q is a + ft therefore if
we assume that a = 2 and a + ft = 4, then ft = 2.

2a + ft = 2(2) + (2) = 6 (9.7)

Step 5. When point P moves closer to point Q, ft gets smaller. This means that the average gradient also gets
smaller. When the point Q overlaps with the point P ft = and the average gradient is given by 2a.

92 CHAPTER 9. GRADIENT AT A POINT

Chapter 10

Linear Programming 1

10.1 Introduction

In everyday life people are interested in knowing the most efficient way of carrying out a task or achieving
a goal. For example, a farmer might want to know how many crops to plant during a season in order to
maximise yield (produce) or a stock broker might want to know how much to invest in stocks in order to
maximise profit. These are examples of optimisation problems, where by optimising we mean finding the
maxima or minima of a function.

You will see optimisation problems of one variable in Grade 12, where there were no restrictions to
the answer. You will be required to find the highest (maximum) or lowest (minimum) possible value of
some function. In this chapter we look at optimisation problems with two variables and where the possible
solutions are restricted.

10.2 Terminology

There are some basic terms which you need to become familiar with for the linear programming chapters.

10.2.1 Decision Variables

The aim of an optimisation problem is to find the values of the decision variables. These values are unknown
at the beginning of the problem. Decision variables usually represent things that can be changed, for example
the rate at which water is consumed or the number of birds living in a certain park.

10.2.2 Objective Function

The objective function is a mathematical combination of the decision variables and represents the function
that we want to optimise (i.e. maximise or minimise). We will only be looking at objective functions which
are functions of two variables. For example, in the case of the farmer, the objective function is the yield
and it is dependent on the amount of crops planted. If the farmer has two crops then the objective function
/ (x, y) is the yield, where x represents the amount of the first crop planted and y represents the amount of
the second crop planted. For the stock broker, assuming that there are two stocks to invest in, the objective
function / (x, y) is the amount of profit earned by investing x rand in the first stock and y rand in the second.

lr This content is available online at <http://siyavula.cnx.Org/content/m30862/l.3/>.

93

94 CHAPTER 10. LINEAR PROGRAMMING

10.2,3 Constraints

Constraints, or restrictions, are often placed on the variables being optimised. For the example of the
farmer, he cannot plant a negative number of crops, therefore the constraints would be:

x>0 /

(10.1)
2/>0.

Other constraints might be that the farmer cannot plant more of the second crop than the first crop and
that no more than 20 units of the first crop can be planted. These constraints can be written as:

x > y /

(10.2)
x< 20

Constraints that have the form

ax + by<c (10.3)

or

ax + by = c (10-4)

are called linear constraints. Examples of linear constraints are:

x + y <
-2x = 7 (10.5)

y<V2

10.2,4 Feasible Region and Points

Constraints mean that we cannot just take any x and y when looking for the x and y that optimise our
objective function. If we think of the variables x and y as a point (x, y) in the xy-plane then we call the
set of all points in the xy-p\&ne that satisfy our constraints the feasible region. Any point in the feasible
region is called a feasible point.
For example, the constraints

x >
2/>0.
mean that only values of x and y that are positive are allowed. Similarly, the constraint

(10.6)

x > y (10.7)

means that only values of x that are greater than or equal to the y values are allowed.

x < 20 (10.8)

means that only x values which are less than or equal to 20 are allowed.

tip: The constraints are used to create bounds of the solution.

95

10.2.5 The Solution

tip: Points that satisfy the constraints are called feasible solutions.

Once we have determined the feasible region the solution of our problem will be the feasible point where the
objective function is a maximum / minimum. Sometimes there will be more than one feasible point where
the objective function is a maximum/minimum — in this case we have more than one solution.

10.3 Example of a Problem

A simple problem that can be solved with linear programming involves Mrs Nkosi and her farm.

Mrs Nkosi grows mielies and potatoes on a farm of 100 m 2 . She has accepted orders that will need her
to grow at least 40 m 2 of mielies and at least 30 m 2 of potatoes. Market research shows that the demand
this year will be at least twice as much for mielies as for potatoes and so she wants to use at least twice as
much area for mielies as for potatoes. She expects to make a profit of R650 per m 2 for her mielies and Rl
500 per m 2 on her potatoes. How should she divide her land so that she can earn the most profit?

Let m represent the area of mielies grown and let p be the area of potatoes grown.

We shall see how we can solve this problem.

10.4 Method of Linear Programming

10.4.1 Method: Linear Programming

1. Identify the decision variables in the problem.

2. Write constraint equations

3. Write objective function as an equation

4. Solve the problem

10.5 Skills you will need

10.5.1 Writing Constraint Equations

You will need to be comfortable with converting a word description to a mathematical description for linear
programming. Some of the words that are used is summarised in Table 10.1.

Words

Mathematical description

x equals a

x = a

x is greater than a

x > a

x is greater than or equal to a

x > a

x is less than a

x < a

x is less than or equal to a

x < a

x must be at least a

x > a

x must be at most o

x < a

Table 10.1: Phrases and mathematical equivalents.

96 CHAPTER 10. LINEAR PROGRAMMING

Exercise 10.1: Writing constraints as equations (Solution on p. 101.)

Mrs Nkosi grows mielies and potatoes on a farm of 100 m 2 . She has accepted orders that will need
her to grow at least 40 m 2 of mielies and at least 30 m 2 of potatoes. Market research shows that
the demand this year will be at least twice as much for mielies as for potatoes and so she wants to
use at least twice as much area for mielies as for potatoes.

10.5.1.1 constraints as equation

Write the following constraints as equations:

1. Michael is registering for courses at university. Michael needs to register for at least 4 courses.

2. Joyce is also registering for courses at university. She cannot register for more than 7 courses.

3. In a geography test, Simon is allowed to choose 4 questions from each section.

4. A baker can bake at most 50 chocolate cakes in 1 day.

5. Megan and Katja can carry at most 400 koeksisters.

10.5.2 Writing the Objective Function

If the objective function is not given to you as an equation, you will need to be able to convert a word
description to an equation to get the objective function.
You will need to look for words like:

• most profit

• least cost

• largest area

Exercise 10.2: Writing the objective function (Solution on p. 101.)

The cost of hiring a small trailer is R500 per day and the cost of hiring a big trailer is R800 per
day. Write down the objective function that can be used to find the cheapest cost for hiring trailers
for 1 day.

Exercise 10.3: Writing the objective function (Solution on p. 101.)

Mrs Nkosi expects to make a profit of R650 per m 2 for her mielies and Rl 500 per m 2 on her
potatoes. How should she divide her land so that she can earn the most profit?

10.5.2.1 Writing the objective function

1. The EduFurn furniture factory manufactures school chairs and school desks. They make a profit of
R50 on each chair sold and of R60 on each desk sold. Write an equation that will show how much
profit they will make by selling the chairs and desks.

2. A manufacturer makes small screen GPS's and wide screen GPS's. If the profit on small screen GPS's
is R500 and the profit on wide screen GPS's is R250, write an equation that will show the possible
maximum profit.

10.5.3 Solving the Problem

The numerical method involves using the points along the boundary of the feasible region, and determining
which point optimises the objective function.

97

10.5.3.1 Investigation : Numerical Method

Use the objective function

650 x m + 1500 x p
to calculate Mrs Nkosi's profit for the following feasible solutions:

(10.9)

m

P

Profit

60

30

65

30

70

30

66|

33±

Table 10.2

The question is How do you find the feasible region? We will use the graphical method of solving a
system of linear equations to determine the feasible region. We draw all constraints as graphs and mark the
area that satisfies all constraints. This is shown in Figure 10.1 for Mrs Nkosi's farm.

Image not finished

Figure 10.1: Graph of the feasible region

Vertices (singular: vertex) are the points on the graph where two or more of the constraints overlap or
cross. If the linear objective function has a minimum or maximum value, it will occur at one or more of the
vertices of the feasible region.

Now we can use the methods we learnt previously to find the points at the vertices of the feasible region.
In Figure 10.1, vertex A is at the intersection of p = 30 and m = 2p. Therefore, the coordinates of A are
(30,60). Similarly vertex B is at the intersection of p = 30 and m = 100 — p. Therefore the coordinates of
B are (30,70). Vertex C is at the intersection of m = 100 — p and m = 2p, which gives (33|,66|) for the
coordinates of C.

If we now substitute these points into the objective function, we get the following:

m

p

Profit

60

30

81 000

70

30

87 000

66|

33|

89 997

Table 10.3

Therefore Mrs Nkosi makes the most profit if she plants 66 1 m 2 of mielies and 33 1 m 2 of potatoes. Her
profit is R89 997.

Exercise 10.4: Prizes! (Solution on p. 101.)

As part of their opening specials, a furniture store has promised to give away at least 40 prizes
with a total value of at least R2 000. The prizes are kettles and toasters.

98 CHAPTER 10. LINEAR PROGRAMMING

1. If the company decides that there will be at least 10 of each prize, write down two more
inequalities from these constraints.

2. If the cost of manufacturing a kettle is R60 and a toaster is R50, write down an objective
function C which can be used to determine the cost to the company of both kettles and
toasters.

3. Sketch the graph of the feasibility region that can be used to determine all the possible
combinations of kettles and toasters that honour the promises of the company.

4. How many of each prize will represent the cheapest option for the company?

5. How much will this combination of kettles and toasters cost?

10.6 End of Chapter Exercises

1. You are given a test consisting of two sections. The first section is on Algebra and the second section
is on Geometry. You are not allowed to answer more than 10 questions from any section, but you have
to answer at least 4 Algebra questions. The time allowed is not more than 30 minutes. An Algebra
problem will take 2 minutes and a Geometry problem will take 3 minutes each to solve. If you answer
xa Algebra questions and ya Geometry questions,

a. Formulate the constraints which satisfy the above constraints.

b. The Algebra questions carry 5 marks each and the Geometry questions carry 10 marks each. If
T is the total marks, write down an expression for T.

2. A local clinic wants to produce a guide to healthy living. The clinic intends to produce the guide in
two formats: a short video and a printed book. The clinic needs to decide how many of each format to
produce for sale. Estimates show that no more than 10 000 copies of both items together will be sold.
At least 4 000 copies of the video and at least 2 000 copies of the book could be sold, although sales
of the book are not expected to exceed 4 000 copies. Let x v be the number of videos sold, and yt the
number of printed books sold.

a. Write down the constraint inequalities that can be deduced from the given information.

b. Represent these inequalities graphically and indicate the feasible region clearly.

c. The clinic is seeking to maximise the income, I, earned from the sales of the two products. Each
video will sell for R50 and each book for R30. Write down the objective function for the income.

d. What maximum income will be generated by the two guides?

3. A patient in a hospital needs at least 18 grams of protein, 0,006 grams of vitamin C and 0,005 grams
of iron per meal, which consists of two types of food, A and B. Type A contains 9 grams of protein,
0,002 grams of vitamin C and no iron per serving. Type B contains 3 grams of protein, 0,002 grams of
vitamin C and 0,005 grams of iron per serving. The energy value of A is 800 kilojoules and of B 400
kilojoules per serving. A patient is not allowed to have more than 4 servings of A and 5 servings of B.
There are x a servings of A and ys servings of B on the patient's plate.

a. Write down in terms of xa and ys

1. The mathematical constraints which must be satisfied.

2. The kilojoule intake per meal.

b. Represent the constraints graphically on graph paper. Use the scale 1 unit = 20mm on both axes.

c. Deduce from the graphs, the values of xa and ys which will give the minimum kilojoule intake
per meal for the patient.

4. A certain motorcycle manufacturer produces two basic models, the 'Super X' and the 'Super Y'. These
motorcycles are sold to dealers at a profit of R20 000 per 'Super X' and R10 000 per 'Super Y'. A 'Super
X' requires 150 hours for assembly, 50 hours for painting and finishing and 10 hours for checking and
testing. The 'Super Y' requires 60 hours for assembly, 40 hours for painting and finishing and 20 hours

99

for checking and testing. The total number of hours available per month is: 30 000 in the assembly
department, 13 000 in the painting and finishing department and 5 000 in the checking and testing
department. The above information can be summarised by the following table:

Department

Hours for 'Super X'

Hours for Super 'Y'

Maximum hours avail-
able per month

Assembley

150

60

30 000

Painting and Finishing

50

40

13 000

Checking and Testing

10

20

5 000

Table 10.4

Let x be the number of 'Super X' and y be the number of 'Super Y' models manufactured per month.

a. Write down the set of constraint inequalities.

b. Use the graph paper provided to represent the constraint inequalities.

c. Shade the feasible region on the graph paper.

d. Write down the profit generated in terms of x and y.

e. How many motorcycles of each model must be produced in order to maximise the monthly profit?

f. What is the maximum monthly profit?

5. A group of students plan to sell x hamburgers and y chicken burgers at a rugby match. They have
meat for at most 300 hamburgers and at most 400 chicken burgers. Each burger of both types is sold
in a packet. There are 500 packets available. The demand is likely to be such that the number of
chicken burgers sold is at least half the number of hamburgers sold.

a. Write the constraint inequalities.

b. Two constraint inequalities are shown on the graph paper provided. Represent the remaining
constraint inequalities on the graph paper.

c. Shade the feasible region on the graph paper.

d. A profit of R3 is made on each hamburger sold and R2 on each chicken burger sold. Write the
equation which represents the total profit, P, in terms of x and y.

e. The objective is to maximise profit. How many, of each type of burger, should be sold to maximise
profit?

6. Fashion-cards is a small company that makes two types of cards, type X and type Y. With the available
labour and material, the company can make not more than 150 cards of type X and not more than
120 cards of type Y per week. Altogether they cannot make more than 200 cards per week. There is
an order for at least 40 type X cards and 10 type Y cards per week. Fashion-cards makes a profit of
R5 for each type X card sold and R10 for each type Y card. Let the number of type X cards be x and
the number of type Y cards be y, manufactured per week.

a. One of the constraint inequalities which represents the restrictions above is x < 150. Write the
other constraint inequalities.

b. Represent the constraints graphically and shade the feasible region.

c. Write the equation that represents the profit P (the objective function), in terms of x and y.

d. Calculate the maximum weekly profit.

7. To meet the requirements of a specialised diet a meal is prepared by mixing two types of cereal, Vuka
and Molo. The mixture must contain x packets of Vuka cereal and y packets of Molo cereal. The meal
requires at least 15 g of protein and at least 72 g of carbohydrates. Each packet of Vuka cereal contains
4 g of protein and 16 g of carbohydrates. Each packet of Molo cereal contains 3 g of protein and 24 g
of carbohydrates. There are at most 5 packets of cereal available. The feasible region is shaded on the
attached graph paper.

100 CHAPTER 10. LINEAR PROGRAMMING

a. Write down the constraint inequalities.

b. If Vuka cereal costs R6 per packet and Molo cereal also costs R6 per packet, use the graph to
determine how many packets of each cereal must be used for the mixture to satisfy the above
constraints in each of the following cases:

1. The total cost is a minimum.

2. The total cost is a maximum (give all possibilities).

Image not finished

Figure 10.2

A bicycle manufacturer makes two different models of bicycles, namely mountain bikes and speed bikes.
The bicycle manufacturer works under the following constraints: No more than 5 mountain bicycles
can be assembled daily. No more than 3 speed bicycles can be assembled daily. It takes one man to
assemble a mountain bicycle, two men to assemble a speed bicycle and there are 8 men working at the
bicycle manufacturer. Let x represent the number of mountain bicycles and let y represent the number
of speed bicycles.

a. Determine algebraically the constraints that apply to this problem.

b. Represent the constraints graphically on the graph paper.

c. By means of shading, clearly indicate the feasible region on the graph.

d. The profit on a mountain bicycle is R200 and the profit on a speed bicycle is R600. Write down
an expression to represent the profit on the bicycles.

e. Determine the number of each model bicycle that would maximise the profit to the manufacturer.

101

Solutions to Exercises in Chapter 10

Solution to Exercise 10.1 (p. 96)

Step 1. There are two decision variables: the area used to plant mielies (m) and the area used to plant potatoes

(p).

Step 2. • grow at least 40 m 2 of mielies

• grow at least 30 m 2 of potatoes

• area of farm is 100 m 2

• demand is at least twice as much for mielies as for potatoes
Step 3. • m > 40

• p> 30

• m + p < 100

• m > 2p

Solution to Exercise 10.2 (p. 96)

Step 1. There are two decision variables: the number of big trailers (rib) and the number of small trailers (n s ).
Step 2. The purpose of the objective function is to minimise cost.
Step 3. The cost of hiring n s small trailers for 1 day is:

500 x n s (10.10)

The cost of hiring rib big trailers for 1 day is:

800 x n b (10.11)

Therefore the objective function, which is the total cost of hiring n s small trailers and rib big trailers
for 1 day is:

500 x n s + 800 x n b (10.12)

Solution to Exercise 10.3 (p. 96)

Step 1. There are two decision variables: the area used to plant mielies (m) and the area used to plant potatoes

(P)-
Step 2. The purpose of the objective function is to maximise profit.

Step 3. The profit of planting m m 2 of mielies is:

650 x m (10.13)

The profit of planting p m 2 of potatoes is:

1500 x p (10.14)

Therefore the objective function, which is the total profit of planting mielies and potatoes is:

650 xm+ 1500 xp (10.15)

Solution to Exercise 10.4 (p. 97)

Step 1. Let the number of kettles be Xk and the number of toasters be y t and write down two constraints apart
from Xk > and y t > that must be adhered to.

102 CHAPTER 10. LINEAR PROGRAMMING

Step 2. Since there will be at least 10 of each prize we can write:

x k > 10 (10.16)

and

Vt > 10 (10.17)

Also the store has promised to give away at least 40 prizes in total. Therefore:

x k + Vt > 40 (10.18)

Step 3. The cost of manufacturing a kettle is R60 and a toaster is R50. Therefore the cost the total cost C is:

C = 60a; fe + 50y t (10.19)

step 4 Image not finished

Figure 10.3

Step 5. From the graph, the coordinates of vertex A are (30,10) and the coordinates of vertex B are (10,30).
Step 6. At vertex A, the cost is:

C = 60x k + 50y t

= 60 (30) + 50 (10)

v ' v ' (10.20)

1800 + 500

2300
At vertex B, the cost is:

C = 60x fe + 50y t

= 60 (10) + 50 (30)

(10.21)
600 + 1500

2100
Step 7. The cheapest combination of prizes is 10 kettles and 30 toasters, costing the company R2 100.

Chapter 11

Geometry

11.1 Polygons 1

11.1.1 Introduction

11.1.1.1 Extension : History of Geometry

Work in pairs or groups and investigate the history of the development of geometry in the last 1500 years.
Describe the various stages of development and how different cultures used geometry to improve their lives.
The works of the following people or cultures should be investigated:

1. Islamic geometry (c. 700 - 1500)

a. Thabit ibn Qurra

b. Omar Khayyam

c. Sharafeddin Tusi

2. Geometry in the 17th - 20th centuries (c. 700 - 1500)

11.1.2 Right Pyramids, Right Cones and Spheres

A pyramid is a geometric solid that has a polygon base and the base is joined to a point, called the apex.
Two examples of pyramids are shown in the left-most and centre figures in Figure 11.1. The right-most
figure has an apex which is joined to a circular base and this type of geometric solid is called a cone. Cones
are similar to pyramids except that their bases are circles instead of polygons.

Image not finished

Figure 11.1: Examples of a square pyramid, a triangular pyramid and a cone.

Surface Area of a Pyramid

1 This content is available online at <http://siyavula.cnx.Org/content/m38832/l.l/>.

103

104 CHAPTER 11. GEOMETRY

Khan academy video on solid geometry volumes

Figure 11.2

The surface area of a pyramid is calculated by adding the area of each face together.

Exercise 11.1: Surface Area (Solution on p. 122.)

If a cone has a height of h and a base of radius r, show that the surface area is irr 2 + irry/r 2 + h 2 .

Volume of a Pyramid: The volume of a pyramid is found by:

V= l -A-h (11.1)

where A is the area of the base and h is the height.

A cone is like a pyramid, so the volume of a cone is given by:

1 9 ,

V = -Trr 2 h. 11.2

3 v '

A square pyramid has volume

V=-a 2 h (11.3)

where a is the side length of the square base.

Exercise 11.2: Volume of a Pyramid (Solution on p. 122.)

What is the volume of a square pyramid, 3cm high with a side length of 2cm?

We accept the following formulae for volume and surface area of a sphere (ball) .

Surface area = Airr 2

(11.4)
Volume = ^7rr 3

11.1.2.1 Surface Area and Volume

1. Calculate the volumes and surface areas of the following solids: *Hint for (e): find the perpendicular
height using Pythagoras.

Image not finished

Figure 11.3

2 http:// www.fhsst.org/12D

105

2. Water covers approximately 71% of the Earth's surface. Taking the radius of the Earth to be 6378 km,
what is the total area of land (area not covered by water)?

3. A triangular pyramid is placed on top of a triangular prism. The prism has an equilateral triangle of
side length 20 cm as a base, and has a height of 42 cm. The pyramid has a height of 12 cm.

a. Find the total volume of the object.

b. Find the area of each face of the pyramid.

c. Find the total surface area of the object.

Image not finished

Figure 11.4

11.1.3 Similarity of Polygons

In order for two polygons to be similar the following must be true:

1. All corresponding angles must be congruent.

2. All corresponding sides must be in the same proportion to each other. Refer to the picture below: this
means that the ratio of side AE on the large polygon to the side PT on the small polygon must be the
same as the ratio of side AB to side PQ, BC/QR etc. for all the sides.

Image not finished

Figure 11.5

If

1. A=P; B=Q; C=R; D=S; E=T and

9 AB _ BC _ CD_ _ DE_ _ EA
A - PQ ~ QR ~ RS ~ ST ~ TP

then the polygons ABCDE and PQRST are similar.

Exercise 11.3: Similarity of Polygons (Solution on p. 123.)

Polygons PQTU and PRSU are similar. Find the value of x.

Image not finished

Figure 11.6

3 http://www.fhsst.org/12W
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106 CHAPTER 11. GEOMETRY

11.2 Triangle geometry 5
11.2.1 Triangle Geometry

11.2.1.1 Proportion

Two line segments are divided in the same proportion if the ratios between their parts are equal.

AB _ x _ kx _ DE

B~C ~ y ~ ky ~ E~F ^ ' '

.•.the line segments are in the same proportion (11-6)

Image notjlnished

Figure 11.7

If the line segments are proportional, the following also hold

i C_B _ FE_

x - AC ~ DF

2. AC ■ FE = CB ■ DF

o AB _ DE a , BC _ FE

°" BC ~ FE clllu AB ~ DE

a AB _ D_E anr] AC _ DF

*■ AC ~ DF cllm AB ~ DE

11.2.1.1.1 Proportionality of triangles

Triangles with equal heights have areas which are in the same proportion to each other as the bases of the
triangles.

hi = h 2

. area [U+25B5]ABC _ |BCxfei _ bc *- ' '

• ' area [U+25B5] DEF ' ±EFxh 2 EF

Image not finished

Figure 11.8

5 This content is available online at <http://siyavula.cnx.Org/content/m38833/l.l/>.

107

• A special case of this happens when the bases of the triangles are equal: Triangles with equal bases
between the same parallel lines have the same area.

area [U+25B5] ABC = - ■ h ■ BC = area [U+25B5] DBC (11.8)

Image not finished

Figure 11.9

• Triangles on the same side of the same base, with equal areas, lie between parallel lines.

If area [U+25B5] ABC = area [U+25B5] BDC, (11.9)

Image not finished

Figure 11.10

Theorem 1 Proportion Theorem: A line drawn parallel to one side of a triangle divides the other two sides
proportionally.

Image not finished

Figure 11.11

Given: [U+25B5] ABC with line DE || BC
R.T.P.:

DB = EC (1LU)

Proof: Draw hi from E perpendicular to AD, and hi from D perpendicular to AE.

108 CHAPTER 11. GEOMETRY

Draw BE and CD.

area [U+25B5]BDE

\DB-h x ~ DB

area [U+25B5JCED

\AE-hi AE

\EC-h-2 EC

but area [U+25B5]BDE =

area [U+25B5] CED (equal base and height)

• ' area [U+25B51BDE

area [U+2BBB]CED

■ ■ DB ~

AE

EC

(11.12)

.•. DE divides AB and AC proportionally.
Similarly,

AB ' AC (nl3)

AB AC v '

BD ~ CE

Following from Theorem "Proportion" (Section 11.2.1.1: Proportion), we can prove the midpoint theorem.

Theorem 2 Midpoint Theorem: A line joining the midpoints of two sides of a triangle is parallel to the
third side and equal to half the length of the third side.

Proof: This is a special case of the Proportionality Theorem (Theorem "Proportion" (Section 11.2.1.1:
Proportion)). If AB = BD and AC = AE, and AD = AB + BD = 2AB AE = AC + CB = 2AC then DE ||
BC and BC = 2DE.

Image not finished

Figure 11.12

Theorem 3 Similarity Theorem 1: Equiangular triangles have their sides in proportion and are therefore
similar.

Image not finished

Figure 11.13

Given: [U+25B5] ABC and [U+25B5]DEF with A=D; B=E; C=F
R.T.P.:

AB AC ,

= 11-14

DE DF y '

Construct: G on AB, so that AG = DE, H on AC, so that AH = DF

Proof: In [U+25B5] 's AGH and DEF

109

AG

=

DE

AH

=

D

A

=

D

[U+25B5] AGH

=

[U+25B5]DEF

.-. AGH

=

E=B

.: GH

II

BC

. AG
• ■ AB

=

AH
AC

. DE
• ' AB

=

DF
AC

[U+25B5]ABC

III

[U+25B5] DEF

(const.)
(const.)

(given)

(SAS)

(corres. Z's equal)
(proportion theorem)
(AG = DE; AH = DF)

(11.15)

tip: HI means "is similar to"
Theorem 4 Similarity Theorem 2: Triangles with sides in proportion are equiangular and therefore similar.

Image not finished

Figure 11.14

Given: [U+25B5] ABC with line DE such that

D~B~ EC
R.T.P..DE || BC; [U+25B5]ADE ||| [U+25B5] ABC

Proof: Draw hi from E perpendicular to AD, and /12 from D perpendicular to AE.
Draw BE and CD.

area [U+25B51BDE

area [U+2BB5]CED

hut AR

out DB

• ' area [U+25B51BDE

area [U+25B5]BDE

:.DE\\ BC

and A E D

\DB-hi DB

\AE-h-2 _ AE
EC

\EC-h 2

m (s iven )

area [U+25B5JCED

area [U+25B5]CED

(same side of equal base DE, same area)

A B C (corres Z's)

AC B

(11.16)

(11.17)

[U+25B5] ADE HI [U+25B5] ABC (AAA)

(11.18)
(11.19)

110

CHAPTER 11. GEOMETRY

Theorem 5 Pythagoras' Theorem: The square on the hypotenuse of a right angled triangle is equal to the
sum of the squares on the other two sides.

Given: [U+25B5] ABC with A= 90°

Image not finished

Figure 11.15

Required to prove:BC 2 = AB 2 + AC 2
Proof:

Let C =

X

DAC =

90°-

- x (Z's of a [U+25B5])

DAB =

X

ABD =

90°-

- x (Z's of a [U+25B5])

BDA =

C D A =A= 90°

(11.20)

[U+25B5] ABD 1 1 1 [U+25B5] CB A and [U+25B5] CAD 1 1 1 [U+25B5] CB A (AAA)

■'■~CB~B~A~ \Ca) an CB ~ ~CA ~ \B~Aj

:. AB 2 = CB x BD and AC 2 = CB x CD

(11.21)

(11.22)
(11.23)

AB 2 + AC 2

i.e.BC 2

CB {BD + CD)

CB (CB)

CB 2

AB 2 + AC 2

(11.24)

Exercise 11.4: Triangle Geometry 1

In [U+25B5] GHI, GH || LJ; GJ || LK and j£ = §. Determine ff .

Image not finished

Figure 11.16

(Solution on p. 123.)

Exercise 11.5: Triangle Geometry 2

PQRS is a trapezium, with PQ || RS. Prove that PT • TR = ST • TQ.

(Solution on p. 124.)

Ill

Image not finished

Figure 11.17

11.2.1.1.1.1 Triangle Geometry

1. Calculate SV

Image not finished

Figure 11.18

2. OB = 3. Find VS.

Image not finished

Figure 11.19

3. Given the following figure with the following lengths, find AE, EC and BE. BC = 15 cm, AB = 4 cm,
CD = 18 cm, and ED = 9 cm.

Image not finished

Figure 11.20

4. Using the following figure and lengths, find IJ and KJ. HI = 26 m, KL = 13 m, JL = 9 m and HJ
32 m.

Image not finished

Figure 11.21

112 CHAPTER 11. GEOMETRY

5. Find FH in the following figure.

Image not finished

Figure 11.22

6. BF = 25 m, AB = 13 m, AD = 9 m, DF = 18m. Calculate the lengths of BC, CF, CD, CE and EF,
and find the ratio ^1.

Image not finished

Figure 11.23

7. If LM || JK, calculate y.

Image not finished

Figure 11.24

11.3 Co-ordinate geometry 6
11.3.1 Co-ordinate Geometry

11.3.1.1 Equation of a Line between Two Points

Khan academy video on point-slope and standard form

Figure 11.25

There are many different methods of specifying the requirements for determining the equation of a straight
line. One option is to find the equation of a straight line, when two points are given.

6 This content is available online at <http://siyavula.cnx.Org/content/m38835/l.l/>.

113

Assume that the two points are (x\; j/i) and {x2\ 2/2), and we know that the general form of the equation
for a straight line is:

y = mx + c (11.25)

So, to determine the equation of the line passing through our two points, we need to determine values for
m (the gradient of the line) and c (the y-intercept of the line). The resulting equation is

y — yi = m (x — Xi) (11.26)

where (x\\ 2/1) are the co-ordinates of either given point.

11.3.1.1.1 Finding the second equation for a straight line

This is an example of a set of simultaneous equations, because we can write:

2/i = mx\ + c
y 2 = mx 2 + c
We now have two equations, with two unknowns, m and c.

(11.27)

2/2 - 2/1 = mx 2 - mxi

m = V2-yi

X2—X1

2/i = mil + c
c = 2/1 — rnxi
Now, to make things a bit easier to remember, substitute into (11.25):

(11.28)

y = mx + c

= mx + (2/1 — mx\) (11.29)

y-yi = m(x-xx)

tip: If you are asked to calculate the equation of a line passing through two points, use:

2/2 - 2/1

m =

x 2 - Xi

to calculate m and then use:

(11.30)

y-yi = m(x-xi) (11.31)

to determine the equation.

For example, the equation of the straight line passing through (—1; 1) and (2; 2) is given by first calculating
m

m

X2—xi

(11.32)

112

-Vl

X 2

-x x

2

-1

2-

(-1)

1
:!

114 CHAPTER 11. GEOMETRY

and then substituting this value into

to obtain

Then substitute (— 1; 1) to obtain

S/ — 2/i

m (x — X\)

y-yi

=

I {x-xi).

y-(i)

=

\{x -{-!))

y-i

=

3 X + 3

y

=

lx + l + 1

y

=

3 X + 3

(11.33)
(11.34)

(11.35)

So, j/ = |x + | passes through (— 1; 1) and (2; 2)

Image not finished

Figure 11.26

Exercise 11.6: Equation of Straight Line (Solution on p. 124.)

Find the equation of the straight line passing through (—3; 2) and (5; 8).

11.3.1.2 Equation of a Line through One Point and Parallel or Perpendicular to Another Line

Another method of determining the equation of a straight-line is to be given one point, (x\;yi), and to
be told that the line is parallel or perpendicular to another line. If the equation of the unknown line is
y = mx + c and the equation of the second line is y = ttiqx + Cq, then we know the following:

If the lines are parallel, then m = run

(11.36)
If the lines are perpendicular, then m x rriQ = —1

Once we have determined a value for m, we can then use the given point together with:

y-yi = m(x-xi) (11.37)

to determine the equation of the line.

For example, find the equation of the line that is parallel to y = 2x — 1 and that passes through (— 1; 1).

First we determine m, the slope of the line we are trying to find. Since the line we are looking for is
parallel to y = 2x — 1,

m = 2 (11.38)

115

The equation is found by substituting m and (— 1; 1) into:

y- yi = m(x-xi)
y -l = 2(»-(-l)

y-i = 2(1 + 1)

(11.39)

y - 1 = 2x + 2

y = 2x + 2 + 1
y = 2x + 3

Image not finished

Figure 11.27: The equation of the line passing through (— 1; 1) and parallel to y — 2x — 1 is y — 2x + 3.
It can be seen that the lines are parallel to each other. You can test this by using your ruler and measuring
the perpendicular distance between the lines at different points.

11.3.1.3 Inclination of a Line

Image not finished

Figure 11.28: (a) A line makes an angle 8 with the x-axis. (b) The angle is dependent on the gradient.
If the gradient of / is m/ and the gradient of g is m g then nif > m g and Of > 6 g .

In Figure 11.28(a), we see that the line makes an angle 9 with the x-axis. This angle is known as the
inclination of the line and it is sometimes interesting to know what the value of 9 is.

Firstly, we note that if the gradient changes, then the value of 9 changes (Figure 11.28(b)), so we suspect
that the inclination of a line is related to the gradient. We know that the gradient is a ratio of a change in
the y-direction to a change in the x-direction.

to = — - (11.40)

As

But, in Figure 11.28(a) we see that

tan9 = -xr-

Ax (11.41)

.•. to = tan9

116 CHAPTER 11. GEOMETRY

For example, to find the inclination of the line y = x, we know m = 1

.". tanO = 1

11.42
:.e = 45°

11.3.1.3.1 Co-ordinate Geometry

1. Find the equations of the following lines

a. through points (— 1;3) and (1;4)

b. through points (7; —3) and (0; 4)

c. parallel to y = \x + 3 passing through (— 1; 3)

d. perpendicular to y = — \x + 3 passing through (—1; 2)

e. perpendicular to 2y + x = 6 passing through the origin

2. Find the inclination of the following lines

a. y = 2x — 3

b. y = \x- 7

c. Ay = 3a; + 8

d. y = — |x + 3 (Hint: if m is negative 6 must be in the second quadrant)

e. 3y + x — 3 =

3. Show that the line y = k for any constant k is parallel to the x-axis. (Hint: Show that the inclination
of this line is 0°.)

4. Show that the line x = k for any constant k is parallel to the y-axis. (Hint: Show that the inclination
of this line is 90°.)

11.4 Transformations 7
11.4.1 Transformations

11.4.1.1 Rotation of a Point

When something is moved around a fixed point, we say that it is rotated about the point. What happens
to the coordinates of a point that is rotated by 90° or 180° around the origin?

11.4.1.1.1 Investigation : Rotation of a Point by 90°

Complete the table, by filling in the coordinates of the points shown in the figure.

7 This content is available online at <http://siyavula.cnx.Org/content/m38836/l.l/>.

117

Point

^-coordinate

y-coordinate

A

B

C

D

E

F

G

H

Table 11.1

Image notjtnished

Figure 11.29

What do you notice about the ^-coordinates? What do you notice about the ^-coordinates? What would
happen to the coordinates of point A, if it was rotated to the position of point C? What about point B
rotated to the position of D?

11.4.1.1.2 Investigation : Rotation of a Point by 180°

Complete the table, by filling in the coordinates of the points shown in the figure.

Point

^-coordinate

y-coordinate

A

B

C

D

E

F

G

H

Table 11.2

118 CHAPTER 11. GEOMETRY

Image not finished

Figure 11.30

What do you notice about the x-coordinates? What do you notice about the y-coordinates? What would
happen to the coordinates of point A, if it was rotated to the position of point E? What about point F
rotated to the position of B?

From these activities you should have come to the following conclusions:

• 90° clockwise rotation: The image of a point P(x;y) rotated clockwise through 90° around the origin
is P'(y; —x). We write the rotation as (x; y) — > (y; —x).

• 90° anticlockwise rotation: The image of a point P(x; y) rotated anticlockwise through 90° around the
origin is P'(—y;x). We write the rotation as (x;y) — > (—y;x).

• 180° rotation: The image of a point P(x;y) rotated through 180° around the origin is P'(— x; —y). We
write the rotation as [x; y) — > (—x; —y).

Image not finished

Figure 11.31

Image not finished

Figure 11.32

Image not finished

Figure 11.33

11.4.1.1.3 Rotation

1. For each of the following rotations about the origin: (i) Write down the rule, (ii) Draw a diagram
showing the direction of rotation.

a. OA is rotated to OA' with A(4;2) and A'(-2;4)

b. OB is rotated to OB' with B(-2;5) and B'(5;2)

c. OC is rotated to OC' with C(-l;-4) and C'(l;4)

119

2. Copy AXYZ onto squared paper. The co-ordinates are given on the picture.

a. Rotate AXYZ anti-clockwise through an angle of 90° about the origin to give AX'Y'Z'. Give the
co-ordinates of X , Y and Z .

b. Rotate AXYZ through 180° about the origin to give AX'Y'Z". Give the co-ordinates of X ", Y
and Z".

Image not finished

Figure 11.34

11.4.1.2 Enlargement of a Polygon 1

When something is made larger, we say that it is enlarged. What happens to the coordinates of a polygon
that is enlarged by a factor kl

11.4.1.2.1 Investigation : Enlargement of a Polygon

Complete the table, by filling in the coordinates of the points shown in the figure. Assume each small square
on the plot is 1 unit.

Point

^-coordinate

y-coordinate

A

B

C

D

E

F

G

H

Table 11.3

Image not finished

Figure 11.35

What do you notice about the ^-coordinates? What do you notice about the y-coordinates? What would
happen to the coordinates of point A, if the square ABCD was enlarged by a factor 2?

120

11.4.1.2.2 Investigation : Enlargement of a Polygon 2

CHAPTER 11. GEOMETRY

Image not finished

Figure 11.36

In the figure quadrilateral HIJK has been enlarged by a factor of 2 through the origin to become H'I'J'K'
Complete the following table using the information in the figure.

Co-ordinate

Co-ordinate'

Length

Length'

H=(;)

H' = (;)

OH =

OH' =

1=0)

HH

01 =

or =

J = 0)

J' = (;)

OJ =

OJ' =

K=0)

K' + (;)

OK =

OK' =

Table 11.4

What conclusions can you draw about

1. the co-ordinates

2. the lengths when we enlarge by a factor of 2?

We conclude as follows:

Let the vertices of a triangle have co-ordinates S(a;i;yi), T^JJte), U(cC3;j/3). [U+25B5]S'T'U' is an
enlargement through the origin of [U+25B5]STU by a factor of c (c > 0).

• [U+25B5] STU is a reduction of [U+25B5] S'T'U' by a factor of c.

• [U+25B5] S'T'U' can alternatively be seen as an reduction through the origin of [U+25B5]STU by a
factor of -. (Note that a reduction by - is the same as an enlargement by c).

• The vertices of [U+25B5] S'T'U' are S'(ca;i;cj/i), T'(cx 2 , cy 2 ), U'(caj 3 , cy 3 ).

• The distances from the origin are OS' = cOS, OT' = cOT and OU' = cOU.

Image not finished

Figure 11.37

11.4.1.2.3 Transformations

1. Copy polygon STUV onto squared paper and then answer the following questions.

121

Image not finished

Figure 11.38

a. What are the co-ordinates of polygon STUV?

b. Enlarge the polygon through the origin by a constant factor of c = 2. Draw this on the same grid.
Label it S'T'U'V.

c. What are the co-ordinates of the vertices of S'T'U'V?

2. [U+25B5] ABC is an enlargement of [U+25B5] A'B'C by a constant factor of k through the origin.

a. What are the co-ordinates of the vertices of [U+25B5]ABC and [U+25B5] A'B'C?

b. Giving reasons, calculate the value of k.

c. If the area of [U+25B5]ABC is m times the area of [U+25B5] A'B'C, what is m?

Image not finished

Figure 11.39

3 Image not finished

Figure 11.40

a. What are the co-ordinates of the vertices of polygon MNPQ?

b. Enlarge the polygon through the origin by using a constant factor of c = 3, obtaining polygon
M'N'P'Q'. Draw this on the same set of axes.

c. What are the co-ordinates of the new vertices?

d. Now draw M"N"P"Q" which is an anticlockwise rotation of MNPQ by 90° around the origin.

e. Find the inclination of OM".

122 CHAPTER 11. GEOMETRY

Solutions to Exercises in Chapter 11

Solution to Exercise 11.1 (p. 104)

Image notjtnished

Step 1.

Figure 11.41

Step 2. The cone has two faces: the base and the walls. The base is a circle of radius r and the walls can be
opened out to a sector of a circle.

Image notjtnished

Figure 11.42

This curved surface can be cut into many thin triangles with height close to a (a is called the slant
height). The area of these triangles will add up to | x base x height (of a small triangle) which is
2 x 2irr x a = irra
Step 3. a can be calculated by using the Theorem of Pythagoras. Therefore:

= \fr 2 + h 2 (11.43)

A b = irr 2 (11.44)

TTTTI

(11.45)

irry/r 2 + h 2

A h + A,

7rr 2 + irry/r 2 + h 2

(11.46)

Step 4.
Step 5.

Step 6.

Solution to Exercise 11.2 (p. 104)

Step 1. The volume of a pyramid is

V=-A-h (11.47)

where A is the area of the base and h is the height of the pyramid. For a square base this means

V=-a-a-h (11.48)

where a is the length of the side of the square base.

Image not finished

Figure 11.43

123

Step 2.

Solution to Exercise 11.3 (p. 105)

Step 1. Since the polygons are similar,

Solution to Exercise 11.4 (p. 110)

Step 1.

LI J

J LI
:. [U+25B5] LI J

•2-2-3
1-12
4 cm 3

PQ

PR

=

TU

su

X

=

3

x+(3— x)

4

. X

■ ■ 3

=

3

4

.'. X

=

9

4

G I H

H G I (Corres.Zs)
[U+25B5] GIH (Equiangular [U+25B5] s)

(11.49)

(11.50)

(11.51)

Step 2.

Step 3.

L I K

K LI

[U+25B5] LIK

HJ

.11
and ^f

HJ
31

G I J

J G I (Corres.Zs)
[U+25B5] GIJ (Equiangular [U+25B5] s)

jjf ( [U+25B5] LI J |
j§ ( [U+25B5] LIK

[U+25B5] GIH)
[U+25B5] GIJ)

We need to calculate -j^y: We were given

HI
KI

JK
Kl

HI JI

JI A KI

(11.52)

(11.53)

(11.54)

So rearranging, we have JK = ^KI And:

JI

ji

Kl

JK-
IKI

KI
-KI

\KI

8

:;

(11.55)

124 CHAPTER 11. GEOMETRY

Using this relation:

- 5 V 8

- 3 X 3

40
9

Solution to Exercise 11.5 (p. 110)

Step 1.

Pi = Si (Alt.Zs)

Qi = i?i (Alt.Zs)

.-. [U+25B5] PTQ HI [U+25B5] STR (Equiangular [U+25B5] s)

Step 2.

£g = Sg ( [U+25B5] PTQ 1 1 1 [U+25B5] STR)
.-. PT-TR = ST ■ TQ

Solution to Exercise 11.6 (p. 114)

Step 1.

(si;yi) = (-3; 2)
(attiitt) = (5; 8)

Step 2.

m = V2 ~ Vl

x 2 —x 1

8-2
- 5-(-3)

_6_

5+3
_ 6

Step 3.

y-yi

—

m[x — x\)

y-(2)

=

!(*-(-3))

y

=

| (* + 3) + 2

=

\x+\ -3 + 2

=

4' X/ ^4^4

=

4 X ^ 4

(11.56)

(11.57)

(11.58)

(11.59)

(11.60)

(11.61)

Step 4. The equation of the straight line that passes through (—3; 2) and (5; 8) is y = %x

11

4~ ' 4 ■

Chapter 12
Trigonometry

12.1 Graphs of trig functions 1

12.1.1 History of Trigonometry

Work in pairs or groups and investigate the history of the development of trigonometry. Describe the various
stages of development and how different cultures used trigonometry to improve their lives.
The works of the following people or cultures can be investigated:

1. Cultures

a. Ancient Egyptians

b. Mesopotamians

c. Ancient Indians of the Indus Valley

2. People

b. Hipparchus (circa 150 BC)

c. Ptolemy (circa 100)

d. Aryabhata (circa 499)

e. Omar Khayyam (1048-1131)

g. Nasir al-Din (13th century)

h. al-Kashi and Ulugh Beg (14th century)
i. Bartholemaeus Pitiscus (1595)

12.1.2 Graphs of Trigonometric Functions

12.1.2.1 Functions of the form y = sin (k9)

In the equation, y = sin(k9), k is a constant and has different effects on the graph of the function. The
general shape of the graph of functions of this form is shown in Figure 12.1 for the function / (9) = sin (29).

1 This content is available online at <http://siyavula.cnx.Org/content/m38866/l.l/>.

125

126 CHAPTER 12. TRIGONOMETRY

Image not finished

Figure 12.1: Graph of / (8) = sin (28) (solid line) and the graph of g (8) — sin (8) (dotted line).

12.1.2.1.1 Functions of the form y = sin(kd)

On the same set of axes, plot the following graphs:

1. a(0) = sin0,58

2. b(8) = sinW

3. c{6) = sinl,56

4. d{0) = sinW

5. e(0) = sin2,56

Use your results to deduce the effect of k.

You should have found that the value of k affects the period or frequency of the graph. Notice that in
the case of the sine graph, the period (length of one wave) is given by ^jr~-

These different properties are summarised in Table 12.1.

k > k <

Image not finished

Figure 12.2

Table 12.1: Table summarising general shapes and positions of graphs of functions of the form

y = sin (kx). The curve y = sin (x) is shown as a dotted line.

12.1.2.1.2 Domain and Range

For / (6) = sin (k9), the domain is {8 : e M} because there is no value of 9 G R for which / (8) is undefined.
The range of / (0) = sin [k8) is {/ [8) : / (8) E [-1, 1]}.

12.1.2.1.3 Intercepts

For functions of the form, y = sin (kd), the details of calculating the intercepts with the y axis are given.
There are many x-intercepts.

127

The y-intercept is calculated by setting 9 = 0:

y = sin(k9)
Vint = sin(0) (12.1)

12.1.2.2 Functions of the form y = cos (k9)

In the equation, y = cos(kO), k is a constant and has different effects on the graph of the function. The
general shape of the graph of functions of this form is shown in Figure 12.4 for the function / (9) = cos (29).

Image notjinished

Figure 12.4: Graph of / (8) = cos (28) (solid line) and the graph of g (8) = cos (8) (dotted line).

12.1.2.2.1 Functions of the form y = cos (kff)
On the same set of axes, plot the following graphs:

1. a (9) = cos0,59

2. b(9) = cosl9

3. c(9) =cosl,59

4. d (9) = cos29

5. e(9) = cos2,59

Use your results to deduce the effect of k.

You should have found that the value of k affects the period or frequency of the graph. The period of
the cosine graph is given by ^S~-

These different properties are summarised in Table 12.2.

k > k <

Image notjinished

Figure 12.5

Table 12.2: Table summarising general shapes and positions of graphs of functions of the form
y = cos (kx). The curve y = cos (x) is plotted with a dotted line.

128 CHAPTER 12. TRIGONOMETRY

12.1.2.2.2 Domain and Range

For / (8) = cos (k6), the domain is {8 : 8 e R} because there is no value of 8 G R for which / (8) is undefined.
The range of / {8) = cos (kff) is {/ (6) : / (6) E [-1, 1]}.

12.1.2.2.3 Intercepts

For functions of the form, y = cos (k9), the details of calculating the intercepts with the y axis are given.
The y-intercept is calculated as follows:

y = cos {kff)
y int = cos(O) (12-2)

= 1

12.1.2.3 Functions of the form y = tan(k8)

In the equation, y = tan(kd), k is a constant and has different effects on the graph of the function. The
general shape of the graph of functions of this form is shown in Figure 12.7 for the function / (8) = tan (28).

Image notjtnished

Figure 12.7: The graph of tan (20) (solid line) and the graph of g(6) = tan(8) (dotted line). The
asymptotes are shown as dashed lines.

12.1.2.3.1 Functions of the form y = tan (kff)
On the same set of axes, plot the following graphs:

1. a(8) =tan0,58

2. b(8) = tan\8

3. c(8) =tanl,50

4. d(8) = tan26

5. e (8) = tan2, 56

Use your results to deduce the effect of k.

You should have found that, once again, the value of k affects the periodicity (i.e. frequency) of the
graph. As k increases, the graph is more tightly packed. As k decreases, the graph is more spread out. The
period of the tan graph is given by ^jr-.

These different properties are summarised in Table 12.3.

129

k > k <

Image not finished

Figure 12.8

Table 12.3: Table summarising general shapes and positions of graphs of functions of the form

y = tan {kff).

12.1.2.3.2 Domain and Range

For / (0) = tan{k0), the domain of one branch is {0 : e I — ^-, ^- )} because the function is undefined
The range of / (0) = tan {kff) is {/ {&) : / (0) € (-00, oo)}.

12.1.2.3.3 Intercepts

For functions of the form, y = tan {kff), the details of calculating the intercepts with the x and y axis are
given.

There are many x-intercepts; each one is halfway between the asymptotes.

The y-intercept is calculated as follows:

y = tan [kff)
y int = ton(O) (12-3)

12.1.2.3.4 Asymptotes

The graph of tank0 has asymptotes because as k0 approaches 90°, tank0 approaches infinity. In other words,
there is no defined value of the function at the asymptote values.

12.1.2.4 Functions of the form y = sin [0 + p)

In the equation, y = sin{0+p), p is a constant and has different effects on the graph of the function.
The general shape of the graph of functions of this form is shown in Figure 12.10 for the function / (0) =

sin (0 + 30°).

Image not finished

Figure 12.10: Graph of / (0) = sin (6 + 30°) (solid line) and the graph of g {9) = sin (6) (dotted line).

130

CHAPTER 12. TRIGONOMETRY

12.1.2.4.1 Functions of the Form y = sin (8 + p)

On the same set of axes, plot the following graphs:

a (8) = sin ((
6(6») = sin (I
c (8) = sinQ
d (8) = sin ((
e (8) = sin (<

■90°)

60°)

■90°)
180°

Use your results to deduce the effect of p.

You should have found that the value of p affects the position of the graph along the y-axis (i.e. the
y-intercept) and the position of the graph along the x-axis (i.e. the phase shift). The p value shifts the graph
horizontally. If p is positive, the graph shifts left and if p is negative tha graph shifts right.

These different properties are summarised in Table 12.4.

p>

p <

Image not finished

Figure 12.11

Table 12.4: Table summarising general shapes and positions of graphs of functions of the form
y = sin (8 + p). The curve y = sin (8) is plotted with a dotted line.

12.1.2.4.2 Domain and Range

For f (8) = sin(8 + p), the domain is {8 : e E} because there is no value of 8 G K for which / (8) is
undefined.

The range of / (8) = sin {8 + p) is {/ (9) : / (9) € [-1, 1]}.

12.1.2.4.3 Intercepts

For functions of the form, y = sin (8 + p), the details of calculating the intercept with the y axis are given.
The y-intercept is calculated as follows: set 8 = 0°

y = sin (8 + p)
y int = sin (0 + p)
= sin (p)

(12.4)

131

12.1.2.5 Functions of the form y = cos (0 + p)

In the equation, y = cos(0 + p), p is a constant and has different effects on the graph of the function.
The general shape of the graph of functions of this form is shown in Figure 12.13 for the function / (9) =

cos (0 + 30°).

Image not finished

Figure 12.13: Graph of / (6) = cos (9 + 30°) (solid line) and the graph of g (6) = cos (9) (dotted line).

12.1.2.5.1 Functions of the Form y = cos (9 + p)

On the same set of axes, plot the following graphs:

1. a{0) = cos {9 -90°)

2. b {9) = cos {9- 60°)

3. c{0) = cos9

4. d{9) = cos {9 + 90°)

5. e {9) = cos {9 +180°)

Use your results to deduce the effect of p.

You should have found that the value of p affects the y-intercept and phase shift of the graph. As in the
case of the sine graph, positive values of p shift the cosine graph left while negative p values shift the graph
right.

These different properties are summarised in Table 12.5.

p > p <

Image notjlnished

Figure 12.14

Table 12.5: Table summarising general shapes and positions of graphs of functions of the form
y = cos (9 + p). The curve y = cos9 is plotted with a dotted line.

12.1.2.5.2 Domain and Range

For f (9) = cos(9+p), the domain is {9 : 9 e E} because there is no value of 9 G K for which / (9) is
undefined.

The range of / (0) = cos {0 + p) is {/ (0) : / (0) e [-1, 1]}.

132 CHAPTER 12. TRIGONOMETRY

12.1.2.5.3 Intercepts

For functions of the form, y = cos (9 + p), the details of calculating the intercept with the y axis are given.
The y-intercept is calculated as follows: set 9 = 0°

y = cos (9 + p)
y int = cos(0+p) (12.5)

= cos (p)

12.1.2.6 Functions of the form y = tan (9 + p)

In the equation, y = tan(9 + p), p is a constant and has different effects on the graph of the function.
The general shape of the graph of functions of this form is shown in Figure 12.16 for the function / (ff) =
tan {9 + 30°).

Image not finished

Figure 12.16: The graph of tan (8 + 30°) (solid lines) and the graph of g (6) — tan (9) (dotted lines).

12.1.2.6.1 Functions of the Form y = tan (9 + p)

On the same set of axes, plot the following graphs:

1. a {9) = tan{9- 90°)

2. b{9) = tan{9-60°)

3. c{9) = tan9

4. d (9) = tan {9 + 60°)

5. e{9) = tan{9+ 180°)

Use your results to deduce the effect of p.

You should have found that the value of p once again affects the y-intercept and phase shift of the graph.
There is a horizontal shift to the left if p is positive and to the right if p is negative.

These different properties are summarised in Table 12.6.

k > k <

Image not finished

Figure 12.17

133

Table 12.6: Table summarising general shapes and positions of graphs of functions of the form
y = tan (6 + p). The curve y = tan (0) is plotted with a dotted line.

12.1.2.6.2 Domain and Range

For / (0) = tan (0 + p), the domain for one branch is {0 : s (—90° — p, 90° — p} because the function is
undefined for = -90° - p and = 90° - p.

The range of / (0) = tan (0 + p) is {/ (0) : / (0) € (-00, oo)}.

12.1.2.6.3 Intercepts

For functions of the form, y = tan (0 + p), the details of calculating the intercepts with the y axis are given.
The y-intercept is calculated as follows: set = 0°

y = tan (0 + p)
y mt = tan (p)

(12.6)

12.1.2.6.4 Asymptotes

The graph of tan(8 + p) has asymptotes because as + p approaches 90°, tan(9 + p) approaches infinity.
Thus, there is no defined value of the function at the asymptote values.

12.1.2.6.4.1 Functions of various form

Using your knowledge of the effects of p and k draw a rough sketch of the following graphs without a table
of values.

1. y = sin3x

2. y = —cos2x

3. y = tan\x

4. y = sin (x — 45°)

5. y = cos (x + 45°)

6. y = tan (x — 45°)

7. y = 2sin2x

8. y = sin (x + 30°) + 1

12.2 Trig identities 2

12.2.1 Trigonometric Identities

12.2.1.1 Deriving Values of Trigonometric Functions for 30°, 45° and 60°

Keeping in mind that trigonometric functions apply only to right-angled triangles, we can derive values of
trigonometric functions for 30°, 45° and 60°. We shall start with 45° as this is the easiest.

Take any right-angled triangle with one angle 45°. Then, because one angle is 90°, the third angle is also
45°. So we have an isosceles right-angled triangle as shown in Figure 12.19.

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134 CHAPTER 12. TRIGONOMETRY

Image not finished

Figure 12.19: An isosceles right angled triangle.

If the two equal sides are of length a, then the hypotenuse, h, can be calculated as:

h 2 = a 2 + a 2

= 2a 2 (12.7)

.-. h = V2a

So, we have:

sin (45° 1 = °PP° sitc ( 45 °)

V ' hvDotcnusc

hypotenuse
a

1

v'2

cos ( 45 °) = adjaeent(45°)

hypotenuse
a

1

x/2

oppositc(45°)

tan (A'", ) - °PP° S1IC ^° >

tan [40 ) - adjacGnt (4 5 o)

(12.c

(12.9)

(12.10)

We can try something similar for 30° and 60°. We start with an equilateral triangle and we bisect one angle
as shown in Figure 12.20. This gives us the right-angled triangle that we need, with one angle of 30° and
one angle of 60°.

Image not finished

Figure 12.20: An equilateral triangle with one angle bisected.

If the equal sides are of length a, then the base is |a and the length of the vertical side, v, can be
calculated as:

(12.11)

v 2

=

a 2 -da) 2

=

a 2 - \a 2

=

h 2

V

=

fa

135

So, we have:

sin (3Cn = °PP° sitc ( 30

v / hvDotenusi

hypotenuse

COS (6U ) — hypotenuse

VI

tnr, (1(\°\ — oPPOsitc(30°)

ran^u ) - adjaccnt ( 30 o )

2 a

_ J_

sin(60°) = °PP° sitc ( 60 °)

^ ' hypotenuse

V3

~~ 2

cos (60°) = ad J accnt ( 60 °)

V / hypotenuse

inn (f\C\°\ — QPPOsitc(60°)

zanyou ) — adjacont ( 60 o)
v3„

V3
You do not have to memorise these identities if you know how to work them out.

TIP:

Image not finished

Figure 12.21

12.2.1.2 Alternate Definition for tanO

We know that tanO is defined as: tanO = °^P a ° c s c '*° This can be written as:

tanO = °PP° sitc x hypotenuse

opposite hypotenuse

(12.12)

(12.13)

(12.14)

(12.15)

(12.16)

(12.17)

(12.18)

But, we also know that sinO is defined as: sind = , opp , osltc and that cosO is defined as: cosO = a J accnt

' hypotenuse hypotenuse

136

CHAPTER 12. TRIGONOMETRY

Therefore, we can write

tanO

opposite
hypotenuse

hypotenuse

sinO x —^s

COS&

(12.19)

tip: tan9 can also be defined as: tan9 = §ID 4

cost)

12.2.1.3 A Trigonometric Identity

One of the most useful results of the trigonometric functions is that they are related to each other. We have
seen that tanQ can be written in terms of sinO and cos9. Similarly, we shall show that: sin 2 9 + cos 2 9 = 1
We shall start by considering [U+25B5] ABC,

Image not finished

Figure 12.22

-g^S and cosk

AB
bc-

We see that: sinO

We also know from the Theorem of Pythagoras that: AB 2

So we can write:

AC 2 = BC 2

sin 9 + cos 2

Bcr

BC 2

m 2 +m 2

AC 2 , AB 2

BC 2 "*" BC 2

AC 2 +AB 2

BC 2

(fromPythagoras)
1

(12.20)

Exercise 12.1: Trigonometric Identities A

Simplify using identities:

1. tan 2 9 ■ cos 2 9

2. —km - tan 2 9

(Solution on p. 155.)

Exercise 12.2: Trigonometric Identities B

Prove: ±^2

l + STO

(Solution on p. 155.)

12.2.1.3.1 Trigonometric identities

1. Simplify the following using the fundamental trigonometric identities:
a.

tanQ

b. cos 2 9.tan 2 6 + tan 2 9.sin 2 t

c. 1 - tan 2 9.sin 2 9

137

d. 1 — sin6.cos8.tan8

e. 1 — sin 8

f. (l=sag*\ - cos 2 8

\ cos 2 J

2. Prove the following:

1+sinQ cos6

a.

cosQ I

,2/

b. sin 8 + (cos8 — tan8) (cos8 + tariff) = 1 — tan

(2cos 2 0-l) , ! _ !_ ta „2 e

C

1 T (l+tan 2 0) 1+tan 2

1 cosdta

] 1 _ cosvtan V i

cosO 1
e . 2 a »n8co S fl = y + cos6 ) _ _1

sinu-\-cosv sinu-\-cosu

' cosO | +_/)^ ^„fl _ 1

f - (fif + ton6 *) ■ cos61 :

12.3 Reduction formulae 3
12,3.1 Reduction Formula

Any trigonometric function whose argument is 90° ± 8, 180° ± 8, 270° ± 8 and 360° ± 8 (hence —8) can be
written simply in terms of 8. For example, you may have noticed that the cosine graph is identical to the
sine graph except for a phase shift of 90°. From this we may expect that sin (90° + 8) = cos8.

12.3.1.1 Function Values of 180° ± 8

12.3.1.1.1 Investigation : Reduction Formulae for Function Values of 180° ±8
1. Function Values of (180° - 8)

a. In the figure P and P' lie on the circle with radius 2. OP makes an angle 8 = 30° with the x-axis.

P thus has coordinates (v3; l). If P' is the reflection of P about the y-axis (or the line x = 0),
use symmetry to write down the coordinates of P'.

b. Write down values for sin8, cos8 and tand.

c. Using the coordinates for P' determine sin (180° — 8), cos (180° — 8) and tan (180° — 8).

Image not finished

Figure 12.23

a. From your results try and determine a relationship between the function values of (180° — 8) and
8.

2. Function values of (180° + 8)

a. In the figure P and P' lie on the circle with radius 2. OP makes an angle 8 = 30° with the ir-axis.
P thus has coordinates (\/3; l). P' is the inversion of P through the origin (reflection about both
the x- and y-axes) and lies at an angle of 180° + 8 with the x-axis. Write down the coordinates
ofP'.

b. Using the coordinates for P' determine sin (180° + 8), cos (180° + 8) and tan (180° + 8).

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138 CHAPTER 12. TRIGONOMETRY

c. From your results try and determine a relationship between the function values of (180° + 9) and

e.

Image notjtnished

Figure 12.24

12.3.1.1.2 Investigation : Reduction Formulae for Function Values of 360° ±9

1. Function values of (360° - 9)

a. In the figure P and P' lie on the circle with radius 2. OP makes an angle = 30° with the a;-axis.
P thus has coordinates (V3; l). P' is the reflection of P about the x-axis or the line y = 0. Using
symmetry, write down the coordinates of P'.

b. Using the coordinates for P' determine sin (360° — 9), cos (360° — 9) and tan (360° — 9).

c. From your results try and determine a relationship between the function values of (360° — 9) and

Image notjtnished

Figure 12.25

It is possible to have an angle which is larger than 360°. The angle completes one revolution to give 360°
and then continues to give the required angle. We get the following results:

(12.21)

sin (360° + 9) =

sin9

cos (360° + 9) =

cos9

tan (360° + 9) =

tan9

also, that if k is any integer, then

sin (k360°+9) =

sin9

cos (fc360° + 9) =

cos9

tan (fc360° + 9) =

tan9

(12.22)

Exercise 12.3: Basic use of a reduction formula (Solution on p. 155.)

Write sm293° as the function of an acute angle.

Exercise 12.4: More complicated... (Solution on p. 155.)

Evaluate without using a calculator:

tan 2 2W° - (1 + cosl20°) sin 2 225° (12.23)

139

12.3.1.1.3 Reduction Formulae

1. Write these equations as a function of 9 only:

a. sin (180° -9)

b. cos (180° - 6)

c. cos (360° - 9)

d. cos (360° + 9)

e. tan (180° -9)

f. cos (360° + 6»)

2. Write the following trig functions as a function of an acute angle:

a. sml63°

b. cos327°

c. tan248°

d. cos213°

3. Determine the following without the use of a calculator:

a. tan 150°. sin 30° + cos 330°

b. tan 300°. cos 120°

c. (1 - cos 30°) (1 - sin 210°)

d. cos 780° + sin 315° .tan 420°

4. Determine the following by reducing to an acute angle and using special angles. Do not use a calculator:

a.

cos 300°

b.

sin 135°

c.

cos 150°

d.

tan 330°

0.

sin 120°

f.

tan 2 225°

g-

cos 315°

h.

sm 2 420°

i.

tan 405°

J-

cos 1020°

k.

tan 2 135°

1.

1 - sin 2 210°

12.3.1.2 Function Values of {-6)

When the argument of a trigonometric function is (— 9) we can add 360° without changing the result. Thus
for sine and cosine

sin (-6) = sin (360° -9) = -sin6 (12.24)

cos {-6) = cos (360° - 9) = cos9 (12.25)

12.3.1.3 Function Values of 90° ± 9

12.3.1.3.1 Investigation : Reduction Formulae for Function Values of 90° ± 6
1. Function values of (90° - 9)

140 CHAPTER 12. TRIGONOMETRY

a. In the figure P and P' lie on the circle with radius 2. OP makes an angle 8 = 30° with the a;-axis.
P thus has coordinates (V3; l). P' is the reflection of P about the line y = x. Using symmetry,
write down the coordinates of P'.

b. Using the coordinates for P' determine sin (90° — 9), cos (90° — 9) and tan (90° — 9).

c. From your results try and determine a relationship between the function values of (90° — 9) and

Image not finished

Figure 12.26

2. Function values of (90° + 6)

a. In the figure P and P' lie on the circle with radius 2. OP makes an angle 6 = 30° with the x-axis.
P thus has coordinates (\/3; l)- P' is the rotation of P through 90°. Using symmetry, write down
the coordinates of P'. (Hint: consider P' as the reflection of P about the line y = x followed by a

b. Using the coordinates for P' determine sin (90° + 6), cos (90° + 9) and tan (90° + 9).

c. From your results try and determine a relationship between the function values of (90° + 9) and
8.

Image not finished

Figure 12.27

Complementary angles are positive acute angles that add up to 90°. e.g. 20° and 70° are complementary
angles.

Sine and cosine are known as co-functions. Two functions are called co-functions if / (A) = g (B)
whenever A+ B = 90° (i.e. A and B are complementary angles). The other trig co-functions are secant and
cosecant, and tangent and cotangent.

The function value of an angle is equal to the co-function of its complement (the co-co rule) .

Thus for sine and cosine we have

sin (90° - 9) = cos6

(12.26)
cos (90° -9) = sin9

Exercise 12.5: Co-co rule (Solution on p. 155.)

Write each of the following in terms of 40° using sin (90° — 9) = cos9 and cos (90° — 9) = sin9.

1. cos50°

2. sm320°

3. cos230°

141

12.3.1.4 Function Values of {9 - 90°)

sin (9 - 90°) = -cosO and cos (9 - 90°) = sin9.
These results may be proved as follows:

sin (9- 90°;

and likewise for cos (9 — 90°) = sin9

12.3.1.5 Summary

The following summary may be made

sin [- (90° - 9)]

-sin (90° -9)
—cos9

(12.27)

second quadrant (180° - 9) or (90° + 9)

first quadrant (9) or (90° - 9)

sin (180° -9) = +sin9

all trig functions are positive

cos (180° - 9) = -cos9

sin (360° + 9) = sin9

tan (180° -9) = -tan9

cos (360° + 9) = cos9

sin (90° + 9) = +cos9

tan (360° + 6) = tan9

cos (90° + 9) = -sin9

sin (90° - 9) = sin9

cos (90° -9) = cos9

sin (180° + 9) = -sin9

sin (360° -9) = -sin9

cos (180° + 9) = -cos9

cos (360° -9) = +cos9

tan (180° + 9) = +tan6

tan (360° -9) = -tan9

Table 12.7

TIP:

1. These reduction formulae hold for any angle 9. For convenience, we usually work with 9 as if

it is acute, i.e. 0° < 9 < 90°.
2. When determining function values of 180° ± 9, 360° ± 9 and —9 the functions never change.
3. When determining function values of 90°±# and #—90° the functions changes to its co-function

(co-co rule).

12.3.1.5.1 Function Values of (270° ± 9)

Angles in the third and fourth quadrants may be written as 270° ± 9 with 9 an acute angle. Similar rules to
the above apply. We get

+ 0)

sin (270° -9) = -cos9

sin (270° + 9) = -cos9

cos (270° -9) = -sinO

cos (270° + 9) = +sin9

Table 12.8

142 CHAPTER 12. TRIGONOMETRY

12.4 Equations 4

12.4.1 Solving Trigonometric Equations

In Grade 10 and 11 we focussed on the solution of algebraic equations and excluded equations that dealt
with trigonometric functions (i.e. sin and cos). In this section, the solution of trigonometric equations will
be discussed.

The methods described in previous Grades also apply here. In most cases, trigonometric identities will be
used to simplify equations, before finding the final solution. The final solution can be found either graphically
or using inverse trigonometric functions.

12.4.1.1 Graphical Solution

As an example, to introduce the methods of solving trigonometric equations, consider

sinO = 0,5 (12.28)

As explained in previous Grades, the solution of Equation (12.28) is obtained by examining the intersecting
points of the graphs of:

V = sine

(12.29)

y = 0,5

Both graphs, for —720° < 6 < 720°, are shown in Figure 12.28 and the intersection points of the graphs are
shown by the dots.

Image not finished

Figure 12.28: Plot of y = sinO and y — 0, 5 showing the points of intersection, hence the solutions to
the equation sin9 = 0, 5.

In the domain for e of —720° < 9 < 720°, there are 8 possible solutions for the equation sin9 = 0,5.
These are 9 = [-690°, -570°, -330°, -210°, 30°, 150°, 390°, 510°]

Exercise 12.6: (Solution on p. 156.)

Find 9, if tane + 0,5 = 1,5, with 0° < 9 < 90°. Determine the solution graphically.

12.4.1.2 Algebraic Solution

The inverse trigonometric functions arcsin, arccos and arctan can also be used to solve trigonometric
equations. These may be shown as second functions on your calculator: sin -1 , cos -1 and tan -1 .
Using inverse trigonometric functions, the equation

sin9 = 0,5 (12.30)

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143

is solved as

sin9 = 0,5

9 = arcsin0,5 (12.31)

30°

Exercise 12.7: (Solution on p. 156.)

Find 9, if tan0 + O, 5 = 1,5, with 0° < 9 < 90°. Determine the solution using inverse trigonometric
functions.

Trigonometric equations often look very simple. Consider solving the equation sin9 = 0, 7. We can take
the inverse sine of both sides to find that 9 = sin^ 1 (0,7). If we put this into a calculator we find that
sin^ 1 (0,7) = 44,42°. This is true, however, it does not tell the whole story.

Image not finished

Figure 12.29: The sine graph. The dotted line represents y = 0, 7. There are four points of intersection
on this interval, thus four solutions to sinO = 0, 7.

As you can see from Figure 12.29, there are four possible angles with a sine of 0,7 between —360° and
360°. If we were to extend the range of the sine graph to infinity we would in fact see that there are an
infinite number of solutions to this equation! This difficulty (which is caused by the periodicity of the sine
function) makes solving trigonometric equations much harder than they may seem to be.

Any problem on trigonometric equations will require two pieces of information to solve. The first is the
equation itself and the second is the range in which your answers must lie. The hard part is making sure
you find all of the possible answers within the range. Your calculator will always give you the smallest
answer (i.e. the one that lies between —90° and 90° for tangent and sine and one between 0° and 180° for
cosine). Bearing this in mind we can already solve trigonometric equations within these ranges.

Exercise 12.8: (Solution on p. 156.)

Find the values of x for which sin (|) = 0, 5 if it is given that x < 90°.

We can, of course, solve trigonometric equations in any range by drawing the graph.

Exercise 12.9: (Solution on p. 156.)

For what values of x does sinx = 0, 5, when —360° < x < 360°?

This method can be time consuming and inexact. We shall now look at how to solve these problems
algebraically.

12.4.1.3 Solution using CAST diagrams

12.4.1.3.1 The Sign of the Trigonometric Function

The first step to finding the trigonometry of any angle is to determine the sign of the ratio for a given angle.
We shall do this for the sine function first and then do the same for the cosine and tangent.

144

CHAPTER 12. TRIGONOMETRY

Image not finished

Figure 12.30: The graph and unit circle showing the sign of the sine function.

In Figure 12.30 we have split the sine graph into four quadrants, each 90° wide. We call them quadrants
because they correspond to the four quadrants of the unit circle. We notice from Figure 12.30 that the sine
graph is positive in the 1 st and 2 nd quadrants and negative in the 3 rd and 4 th . Figure 12.31 shows similar
graphs for cosine and tangent.

Image not finished

Figure 12.31: Graphs showing the sign of the cosine and tangent functions.

All of this can be summed up in two ways. Table 12.9 shows which trigonometric functions are positive
and which are negative in each quadrant.

-i st

2 nd

3 rd

4 th

sin

+VE

+VE

-VE

-VE

cos

+VE

-VE

-VE

+VE

tan

+VE

-VE

+VE

-VE

Table 12.9: The signs of the three basic trigonometric functions in each quadrant.

A more convenient way of writing this is to note that all functions are positive in the 1 st quadrant, only
sine is positive in the 2 nd , only tangent in the 3 rd and only cosine in the 4 th . We express this using the
CAST diagram (Figure 12.32). This diagram is known as a CAST diagram as the letters, taken anticlockwise
from the bottom right, read C-A-S-T. The letter in each quadrant tells us which trigonometric functions are
positive in that quadrant. The 'A' in the 1 st quadrant stands for all (meaning sine, cosine and tangent are
all positive in this quadrant). 'S', 'C and 'T' ,of course, stand for sine, cosine and tangent. The diagram is
shown in two forms. The version on the left shows the CAST diagram including the unit circle. This version
is useful for equations which lie in large or negative ranges. The simpler version on the right is useful for
ranges between 0° and 360°. Another useful diagram shown in Figure 12.32 gives the formulae to use in
each quadrant when solving a trigonometric equation.

145

Image not finished

Figure 12.32: The two forms of the CAST diagram and the formulae in each quadrant.

12.4.1.3.2 Magnitude of the trigonometric functions

Now that we know in which quadrants our solutions lie, we need to know which angles in these quadrants
satisfy our equation.

Calculators give us the smallest possible answer (sometimes negative) which satisfies the equation. For
example, if we wish to solve sinO = 0,3 we can apply the inverse sine function to both sides of the equation
to find:

9 = arcsinO, 3

(12.32)
= 17,46°

However, we know that this is just one of infinitely many possible answers. We get the rest of the answers
by finding relationships between this small angle, 9, and answers in other quadrants. To do this we use our
small angle 9 as a reference angle. We then look at the sign of the trigonometric function in order to
decide in which quadrants we need to work (using the CAST diagram) and add multiples of the period to
each, remembering that sine, cosine and tangent are periodic (repeating) functions. To add multiples of the
period we use 360° • n (where n is an integer) for sine and cosine and 180° • n, n € Z, for the tangent.

Exercise 12.10: (Solution on p. 157.)

Solve for 9:

sin9 = 0,3 (12.33)

12.4.1.4 General Solution Using Periodicity

Up until now we have only solved trigonometric equations where the argument (the bit after the function,
e.g. the 9 in cos9 or the (2x — 7) in tan (2x — 7)), has been 9. If there is anything more complicated than this
we need to be a little more careful. Let us try to solve tan (2x — 10°) = 2, 5 in the range —360° < x < 360°.
We want solutions for positive tangent so using our CAST diagram we know to look in the 1 st and 3 rd
quadrants. Our calculator tells us that arctan(2,5) = 68,2°. This is our reference angle. So to find the
general solution we proceed as follows:

tan {2x- 10°) = 2,5 [68,2°]

I: 2x -10° = 68, 2° + 180° -n

(12.34)
2x = 78, 2° + 180° -n

x = 39,1° + 90° -n,neZ

This is the general solution. Notice that we added the 10° and divided by 2 only at the end. Notice that we
added 180° • n because the tangent has a period of 180°. This is also divided by 2 in the last step to keep

146 CHAPTER 12. TRIGONOMETRY

the equation balanced. We chose quadrants I and III because tan was positive and we used the formulae 8 in
quadrant I and (180° + 8) in quadrant III. To find solutions where —360° < x < 360° we substitue integers
for n:

• n = 0; x = 39,1°; 129,1°

• n =l- x = 129,1°; 219,1°

• n = 2; x = 219,1°; 309,1°

• n = 3; x = 309,1°; 399,1° (too big!)

• n= -1; x = -50,9°; 39,1°

• n= -2; aj = -140,1°; -50,9°

• n = -3; x = -230,9°; -140,9°

• n = -4; x = -320,9°; -230,9°

Solution: x = -320, 9°; -230°; -140, 9°; -50, 9°; 39, 1°; 129, 1°; 219, 1° and 309, 1°

12.4.1.5 Linear Trigonometric Equations

Just like with regular equations without trigonometric functions, solving trigonometric equations can become
a lot more complicated. You should solve these just like normal equations to isolate a single trigonometric
ratio. Then you follow the strategy outlined in the previous section.

Exercise 12.11: (Solution on p. 157.)

Write down the general solution for 3cos (8 — 15°) — 1 = —2, 583

12.4.1.6 Quadratic and Higher Order Trigonometric Equations

The simplest quadratic trigonometric equation is of the form

sin 2 x-2 = -1,5 (12.35)

This type of equation can be easily solved by rearranging to get a more familiar linear equation

sin x = 0, 5

(12.36)

=> sinx = ±-y/0, 5

This gives two linear trigonometric equations. The solutions to either of these equations will satisfy the

The next level of complexity comes when we need to solve a trinomial which contains trigonometric
functions. It is much easier in this case to use temporary variables. Consider solving

tan 2 (2x + 1) + Stan (2x + 1) + 2 = (12.37)

Here we notice that tan (2x + 1) occurs twice in the equation, hence we let y = tan (2x + 1) and rewrite:

y 2 + 3y + 2 = (12.38)

That should look rather more familiar. We can immediately write down the factorised form and the solutions:

(y + l)(y + 2)=0 ,

v Jy ' (12.39)

=>y=-l OR y=-2

Next we just substitute back for the temporary variable: tan (2x + 1) = — 1 or tan (2x + 1) = —2 And
then we are left with two linear trigonometric equations. Be careful: sometimes one of the two solutions
will be outside the range of the trigonometric function. In that case you need to discard that solution. For

147

example consider the same equation with cosines instead of tangents cos 2 (2x + 1) + 3cos (2x + 1) + 2 =
Using the same method we find that cos (2x + 1) = — 1 or cos (2x + 1) = — 2 The second solution cannot
be valid as cosine must lie between —1 and 1. We must, therefore, reject the second equation. Only solutions
to the first equation will be valid.

12.4.1.7 More Complex Trigonometric Equations

Here are two examples on the level of the hardest trigonometric equations you are likely to encounter. They
require using everything that you have learnt in this chapter. If you can solve these, you should be able to
solve anything!

Exercise 12.12: (Solution on p. 158.)

Solve 2cos 2 x - cosx - 1 = for x G [-180°; 360°]

Exercise 12.13: (Solution on p. 158.)

Solve for x in the interval [—360°; 360°]:

2sin x — sinxcosx = (12.40)

12.4.1.7.1 Solving Trigonometric Equations

1. a. Find the general solution of each of the following equations. Give answers to one decimal place,
b. Find all solutions in the interval 9 e [-180°; 360°] .

1. sin9= -0,327

2. cos6 = 0,231

3. tan9 = -1,375

4. sin0 = 2,439

2. a. Find the general solution of each of the following equations. Give answers to one decimal place,
b. Find all solutions in the interval 9 £ [0°; 360°] .

1. cos0 =

2. sinO = ^

3. 2cos0 - \/3 =

4. tan9 = -1

5. 5cos9 = -2

6. 3sin9 =—1,5

7. 2cosd+ 1,3 =

8. O,5£an0 + 2,5 = 1,7

3. a. Write down the general solution for x if tanx = —1, 12.
b. Hence determine values of x € [—180°; 180°].

4. a. Write down the general solution for if sin9 = —0,61.
b. Hence determine values of 9 G [0°; 720°].

5. a. Solve for A if sin (A + 20°) = 0, 53

b. Write down the values of Ae [0°; 360°]

6. a. Solve for x if cos (x + 30°) = 0, 32

b. Write down the values of x e [-180°; 360°]

7. a. Solve for 6 if sin 2 {&) + 0, 5sin0 =

b. Write down the values of 9 e [0°; 360°]

148 CHAPTER 12. TRIGONOMETRY

12.5 Cosine and sine identities 5
12,5.1 Sine and Cosine Identities

There are a few identities relating to the trigonometric functions that make working with triangles easier.
These are:

1. the sine rule

2. the cosine rule

3. the area rule

and will be described and applied in this section.

12.5.1.1 The Sine Rule

Definition 12.1: The Sine Rule

The sine rule applies to any triangle: ^^ = SI f^ = smc wnere a [ s the side opposite A, b is the

side opposite B and c is the side opposite C-
Consider [U+25B5] ABC.

Image not finished

Figure 12.33

The area of [U+25B5] ABC can be written as: area [U+25B5] ABC = \c-h. However, h can be calculated
in terms of A or B as:

sin A = j;

h = b ■ sin A

(12.41)

and

(12.42)

sin B = -

a

h = a ■ sin B

Therefore the area of [U+25B5] ABC is: |c • h = \c ■ b ■ sin A= \c- a ■ sin B

Similarly, by drawing the perpendicular between point B and line AC we can show that: \c-b ■ sin A-
\a- b ■ sin C

Therefore the area of [U+25B5] ABC is: \c-b ■ sin A= \c ■ a ■ sin B= \a ■ b ■ sin C

If we divide through by \a-b- c, we get: si2 ^ = ^^ = m ^-

This is known as the sine rule and applies to any triangle, right angled or not.

5 This content is available online at <http://siyavula.cnx.Org/content/m38873/l.l/>.

149

Exercise 12.14: Lighthouses (Solution on p. 159.)

There is a coastline with two lighthouses, one on either side of a beach. The two lighthouses are
0, 67 km apart and one is exactly due east of the other. The lighthouses tell how close a boat is
by taking bearings to the boat (remember - a bearing is an angle measured clockwise from north).
These bearings are shown. Use the sine rule to calculate how far the boat is from each lighthouse.

Image notjtnished

Figure 12.34

12.5.1.1.1 Sine Rule

1. Show that SJ^A = sirR = str^c is equ i va i ent to: _s_ = _&_ = _^i_ Note: eit her of these two forms

sinA sinB sinC

can be used.

2. Find all the unknown sides and angles of the following triangles:

a. [U+25B5]PQR in which Q= 64°; R= 24° and r = 3

b. [U+25B5] KLM in which K= 43°; M= 50° and m = 1

c. [U+25B5] ABC in which A= 32, 7°; C= 70, 5° and a = 52, 3

d. [U+25B5]XYZ in which X= 56°; Z= 40° and x = 50

3. In [U+25B5] ABC, A= 116°; C= 32° and AC = 23 m. Find the length of the side AB.

4. In [U+25B5]RST, R= 19°; S= 30° and RT = 120 km. Find the length of the side ST.

5. In [U+25B5]KMS, K= 20°; M= 100° and s = 23 cm. Find the length of the side m.

12.5.1.2 The Cosine Rule

Definition 12.2: The Cosine Rule

The cosine rule applies to any triangle and states that:

a 2 = b 2 + c 2 — 2bccos A

b 2 = c 2 + a 2 - 2cacos B (12.43)

c 2 = a 2 + b 2 - 2abcos C

where a is the side opposite A, b is the side opposite B and c is the side opposite C-

The cosine rule relates the length of a side of a triangle to the angle opposite it and the lengths of the
other two sides.

Consider [U+25B5] ABC which we will use to show that: a 2 = b 2 + c 2 — 2bccos A ■

150 CHAPTER 12. TRIGONOMETRY

Image notjtnished

Figure 12.35

In [U+25B5] DCB: a 2 = (c - d) + h 2 from the theorem of Pythagoras.
In [U+25B5] ACD: b 2 = d 2 + h 2 from the theorem of Pythagoras.
We can eliminate h 2 from and to get:

b 2 -d 2 = a 2 -{c-df

a 2 = b 2 + (c 2 - 2cd + d 2 ) - d 2
= b 2 + c 2 - led + d 2 -d 2
b 2 + c 2 - led

(12.44)

In order to eliminate d we look at [U+25B5] ACD, where we have: cos A= i- So, d = bcos A ■ Substituting

this into (12.44), we get: a 2 = b 2 + c 2 - Ibccos A

The other cases can be proved in an identical manner.

Exercise 12.15: (Solution on p. 159.)

Find A:

Image notjtnished

Figure 12.36

12.5.1.2.1 The Cosine Rule

1. Solve the following triangles i.e. find all unknown sides and angles

a. [U+25B5] ABC in which A= 70°; 6 = 4 and c = 9

b. [U+25B5]XYZ in which Y= 112°; x = 2 and y = 3

c. [U+25B5]RST in which RS= 2; ST= 3 and RT= 5

d. [U+25B5] KLM in which KL= 5; LM= 10 and KM= 7

e. [U+25B5] JHK in which H= 130°; JH= 13 and HK= 8

f. [U+25B5] DEF in which d = 4; e = 5 and / = 7

2. Find the length of the third side of the [U+25B5]XYZ where:

a. X= 71, 4°; y = 3, 42 km and z = 4, 03 km

b. ; x = 103, 2 cm; Y= 20, 8° and z = 44, 59 cm

3. Determine the largest angle in:

151

a. [U+25B5] JHK in which JH= 6; HK= 4 and JK= 3

b. [U+25B5] PQR where p = 50; q = 70 and r = 60

12.5.1.3 The Area Rule

Definition 12.3: The Area Rule

The area rule applies to any triangle and states that the area of a triangle is given by half the
product of any two sides with the sine of the angle between them.

That means that in the [U+25B5] DEF, the area is given by: A = \DE ■ EFsin E= \EF ■ FDsin F=
\FD ■ DEsin D

Image not finished

Figure 12.37

In order show that this is true for all triangles, consider [U+25B5] ABC.

Image not finished

Figure 12.38

The area of any triangle is half the product of the base and the perpendicular height. For [U+25B5] ABC,
this is: A = \c-h. However, h can be written in terms of A as: h = bsin A So, the area of [U+25B5] ABC

is: A= \c- bsin A ■

Using an identical method, the area rule can be shown for the other two angles.

Exercise 12.16: The Area Rule (Solution on p. 160.)

Find the area of [U+25B5] ABC:

Image not finished

Figure 12.39

12.5.1.3.1 The Area Rule

Draw sketches of the figures you use in this exercise.

152

CHAPTER 12. TRIGONOMETRY

1. Find the area of [U+25B5]PQR in which:

a. p= 40°; q = 9 and r = 25

b. Q= 30°; r = 10 and p = 7

c. r= 110°; p= 8 and g = 9

2. Find the area of:

a. [U+25B5] XYZ with XY= 6 cm; XZ= 7 cm and Z= 28°

b. [U+25B5]PQR with PR= 52 cm; PQ= 29 cm and P= 58, 9°

c. [U+25B5]EFG with FG= 2,5 cm; EG= 7,9 cm and G= 125°

3. Determine the area of a parallelogram in which two adjacent sides are 10 cm and 13 cm and the angle
between them is 55°.

4. If the area of [U+25B5] ABC is 5000 m 2 with a = 150 m and b = 70 m, what are the two possible sizes

of C?
12,5.2 Summary of the Trigonometric Rules and Identities

Pythagorean Identity

Ratio Identity

cos 2 9 + sin 2 9 = 1

tan6 = sml

cosu

Table 12.10

Odd/Even Identities

Periodicity Identities

Cofunction Identities

sin (—0) = —sin8

sin(0± 360°) = sin9

sin (90° - 6) = cos9

cos (—9)= cos9

cos {9 ±360°) = cos9

cos (90° - 9) = sin9

Sine Rule

Area Rule

Cosine Rule

Area = ^bccosA

a 2 = b 2 + c 2 - IbccosA

sinA sinB sinC

a b c

Area = \d,ccosB

b 2 = a 2 + c 2 — 2accosB

Area = ^abcosC

c 2 = a 2 + b 2 - 2abcosC

Table 12.11

12,5.3 Exercises

1. Q is a ship at a point 10 km due South of another ship P. R is a lighthouse on the coast such that
P=Q= 50°. Determine:

153

a. the distance QR

b. the shortest distance from the lighthouse to the line joining the two ships (PQ).

Image notjtnished

Figure 12.40

2. WXYZ is a trapezium (WX||XZ) with WX= 3 m; YZ= 1,5 m;Z= 120° and W= 30°

a. Determine the distances XZ and XY.

b. Find the angle C-

Image notjtnished

Figure 12.41

3. On a flight from Johannesburg to Cape Town, the pilot discovers that he has been flying 3° off course.
At this point the plane is 500 km from Johannesburg. The direct distance between Cape Town and
Johannesburg airports is 1 552 km. Determine, to the nearest km:

a. The distance the plane has to travel to get to Cape Town and hence the extra distance that the
plane has had to travel due to the pilot's error.

b. The correction, to one hundredth of a degree, to the plane's heading (or direction).

4. ABCD is a trapezium (ie. AB||CD). AB= x; B A D = a; B C D = b and B D C = c. Find an
expression for the length of CD in terms of x, a, 6 and c.

Image notjtnished

Figure 12.42

5. A surveyor is trying to determine the distance between points X and Z. However the distance cannot
be determined directly as a ridge lies between the two points. From a point Y which is equidistant

from X and Z, he measures the angle X Y Z.

a. If XY= x and XY Z = 6, show that XZ= x^2 (1 - cosd).

b. Calculate XZ (to the nearest kilometre) if x = 240 km and 9 = 132°.

154 CHAPTER 12. TRIGONOMETRY

Image not finished

Figure 12.43

6. Find the area of WXYZ (to two decimal places):

Image not finished

Figure 12.44

7. Find the area of the shaded triangle in terms of x, a, (3, 8 and <f>:

Image not finished

Figure 12.45

155

Solutions to Exercises in Chapter 12

Solution to Exercise 12.1 (p. 136)

Step 1.

tan 2 9 ■ cos 2 8

„2/i

cos 2 9

COS

(12.45)

sin

Step 2.

tan 2 e

cos 2 9

1 sin 2 9

cos^v cos^

l-sin 2 9

cos 2 9

cos i

(12.46)

Solution to Exercise 12.2 (p. 136)

Step 1.

LHS

1 — sinx v l-\-sinx
cosx l-\-sinx

cosx(l-\- sinx)

cos 1 x
cosx(l-\-sinx)

(12.47)

l-\-sinx

RHS

Solution to Exercise 12.3 (p. 138)
Step 1.

where we used the fact that sin (360°
are in fact equal:

sin293° = sin (360° -67°)
= — sin67°

) = —sinO. Check, using your calculator, that

sm293°
— sin67°

-0,92- • •
-0,92- ••

(12.48)
these values

(12.49)

Solution to Exercise 12.4 (p. 138)
Step 1.

°\ „„-„2r>r>rO

tan 2 210° - (1 + cosl20°) sin A 225

[tan (180° + 30°)] 2 - [1 + cos (180° - 60°)] • [sin (180° + 45°)] 5
(tan30°f - [1 + (-cos60°)] • (-sm45°) 2

(12.50)

3 (2) (2/

1 _ 1 _ J_
3 4 12

Solution to Exercise 12.5 (p. 140)

156 CHAPTER 12. TRIGONOMETRY

Step 1. a. cos50° = cos (90° - 40°) = sm40°

b. sm320° = sin (360° - 40°) = -sm40°

c. cos230° = cos (180° + 50°) = -cos50° = -cos (90° - 40°) = -sm40°

Solution to Exercise 12.6 (p. 142)

Step 1.

tan9 + 0,5 = 1,5
tan9 = 1

(12.51)

Step 2.

y = tanO

y = i

(12.52)

Image not finished

Step 3.

Figure 12.46

The graphs intersect at = 45°.

Solution to Exercise 12.7 (p. 143)

Step 1.

tan9 + 0,5 =

1,5

tanO =

1

:. 6 =

arctanl

—

45°

(12.53)

Solution to Exercise 12.8 (p. 143)

Step 1. Because we are told that x is an acute angle, we can simply apply an inverse trigonometric function
to both sides.

«»(§)

=

0,5

^ 2

=

arcsinO, 5

^ 2

=

30°

.'. X

=

60°

(12.54)

Solution to Exercise 12.9 (p. 143)

Step 1. We take a look at the graph of sinx = 0,5 on the interval [—360°, 360°]. We want to know when the
y value of the graph is 0, 5, so we draw in a line at y = 0, 5.

Image not finished

Figure 12.47

157

Step 2. Notice that this line touches the graph four times. This means that there are four solutions to the

equation.
Step 3. Read off the x values of those intercepts from the graph as x = —330°, —210°, 30° and 150°.

Image not finished

Figure 12.48

Solution to Exercise 12.10 (p. 145)

Step 1. We look at the sign of the trigonometric function. sinO is given as a positive amount (0, 3). Reference
to the CAST diagram shows that sine is positive in the first and second quadrants.

s

A

T

C

Table 12.12

Step 2. The small angle is the angle returned by the calculator:

sinO

0,3

arcsinO, 3

17,46°

Step 3. Our solution lies in quadrants I and II. We therefore use and 180°
periodicity of sine.

(12.55)

and add the 360° ■ n for the

180° - e

e

180° + e

360° -

Table 12.13

I: 9 = 17,46° + 360° -n 7 neZ

II : 6 = 180° - 17, 46° + 360° -n,neZ
= 162,54° + 360° -n,neZ

(12.56)

This is called the general solution.

Step 4. We can then find all the values of by substituting n = ..., — 1,0, 1, 2, ...etc. For example, If n =
0, = 17,46°; 162,54° If n = 1, 9 = 377,46°; 522,54° If n = -1, = -342,54°; -197,46° We
can find as many as we like or find specific solutions in a given interval by choosing more values for n.

Solution to Exercise 12.11 (p. 146)

158 CHAPTER 12. TRIGONOMETRY

Step 1.

3cos(0- 15°) - 1 = -2,583

3cos(0-15°) = -1,583

cos {9 -15°) = -0,5276...

referenceangle : (6 - 15°) = 58,2°

V ' (12.57)

II: 61-15° = 180° -58,2° + 360° -n,n£ Z

= 136,8° + 360° -n,neZ

III: 0-15° = 180° + 58,2° + 360° -n,ne Z

= 253,2° + 360° -n,neZ

Solution to Exercise 12.12 (p. 147)

Step 1. We note that cosx occurs twice in the equation. So, let y = cosx. Then we have 2y 2 — y — 1 = Note

that with practice you may be able to leave out this step.
Step 2. Factorising yields

(2y+l)(y-l) = (12.58)

.-. y=-0,5 or y = l (12.59)

Step 3. We thus get

cosx = —0,5 or cosx = 1 (12.60)

Both equations are valid (i.e. lie in the range of cosine). General solution:

cosx = -0,5 [60°]

II: x = 180° - 60° + 360° -n,ne Z

120° + 360°-n,neZ (12.61)

III: x = 180° + 60° + 360° -n,n€ Z
= 240° + 360° -n,neZ

cosx = 1 [90°]

I; IV : x = 0° + 360° • n, n e Z (12.62)

= 360°-n,neZ

Now we find the specific solutions in the interval [—180°; 360°]. Appropriate values of n yield

x = -120°; 0°; 120°; 240°; 360° (12.63)

Solution to Exercise 12.13 (p. 147)

Step 1. Factorising yields

sinx (2sinx — cosx) = (12.64)

which gives two equations

sinx = (12.65)

159

2sinx

=

COSX

2sinx
cosx

=

cosx
cosx

2tanx

=

1

tanx

=

1

2

sinx =

[0°]

x =

180°

•n,n6Z

IX =

l

2

[26,57°

(12.66)

Step 2. General solution:

™t - n fn°l

(12.67)

(12.68)

26, 57° + 180° -n,neZ

Specific solution in the interval [-360°; 360°]: x = -360°; -206, 57°; -180°; -26, 57°; 0°; 26, 57°; 180°; 206, 25°; 360°

Solution to Exercise 12.14 (p. 149)

Step 1. We can see that the two lighthouses and the boat form a triangle. Since we know the distance between
the lighthouses and we have two angles we can use trigonometry to find the remaining two sides of the
triangle, the distance of the boat from the two lighthouses.

Image notjtnished

Figure 12.49

Step 2. We need to know the lengths of the two sides AC and BC. We can use the sine rule to find our missing
lengths.

BC

=

AB

sinA

sinC

BC

=

AB-sinA

=

sinC
(0,67km)sin(37°)

sin(128°)

=

0,51km

AC

=

AB

sinB

sinC

AC

=

AB-sinB

=

(0.

sinC
,67km)sm(15°)

sm(128°)

=

0, 22km

(12.69)

(12.70)

Solution to Exercise 12.15 (p. 150)

160 CHAPTER 12. TRIGONOMETRY

Step 1.

(12.71)

a 2

=

b 2 +

c 2 — 2bdcos j

A

—

b 2 +c 2 -a 2
2bc

8 2 +5 2 -7 2
2-8-5

=

0,5

A

=

arccosO, 5 = 60°

Solution to Exercise 12.16 (p. 151)

Step 1. [U+25B5] ABC is isosceles, therefore AB=AC= 7 and C=B= 50°. Hence A= 180° - 50° - 50° = 80°.
Now we can use the area rule to find the area:

A = ^cbsin A

= 5 • 7 • 7 • sin80° (12.72)

24,13

Chapter 13

Statistics

13.1 Standard deviation and variance 1

13.1.1 Introduction

This chapter gives you an opportunity to build on what you have learned in previous grades about data
handling and probability. The work done will be mostly of a practical nature. Through problem solving
and activities, you will end up mastering further methods of collecting, organising, displaying and analysing
data. You will also learn how to interpret data, and not always to accept the data at face value, because
data is sometimes misused and abused in order to try to falsely prove or support a viewpoint. Measures
of central tendency (mean, median and mode) and dispersion (range, percentiles, quartiles, inter-quartile,
semi-inter-quartile range, variance and standard deviation) will be investigated. Of course, the activities
involving probability will be familiar to most of you - for example, you may have played dice or card games
even before you came to school. Your basic understanding of probability and chance gained so far will be
deepened to enable you to come to a better understanding of how chance and uncertainty can be measured
and understood.

13.1.2 Standard Deviation and Variance

The measures of central tendency (mean, median and mode) and measures of dispersion (quartiles, per-
centiles, ranges) provide information on the data values at the centre of the data set and provide information
on the spread of the data. The information on the spread of the data is however based on data values at
specific points in the data set, e.g. the end points for range and data points that divide the data set into 4
equal groups for the quartiles. The behaviour of the entire data set is therefore not examined.

A method of determining the spread of data is by calculating a measure of the possible distances between
the data and the mean. The two important measures that are used are called the variance and the standard
deviation of the data set.

13.1.2.1 Variance

The variance of a data set is the average squared distance between the mean of the data set and each data
value. An example of what this means is shown in Figure 13.1. The graph represents the results of 100 tosses
of a fair coin, which resulted in 45 heads and 55 tails. The mean of the results is 50. The squared distance
between the heads value and the mean is (45 — 50) = 25 and the squared distance between the tails value
and the mean is (55 — 50) = 25. The average of these two squared distances gives the variance, which is
|(25 + 25) = 25.

1 This content is available online at <http://siyavula.cnx.Org/content/m38858/l.l/>.

161

162 CHAPTER 13. STATISTICS

Image notjtnished

Figure 13.1

13.1.2.1.1 Population Variance

Let the population consist of n elements {x\,X2, ...,x n }, with mean x (read as "x bar"). The variance of the
population, denoted by a 2 , is the average of the square of the distance of each data value from the mean
value.

a 2 = (E(*-X))\ (131)

n
Since the population variance is squared, it is not directly comparable with the mean and the data themselves.

13.1.2.1.2 Sample Variance

Let the sample consist of the n elements {x\,X2, ...,x n }, taken from the population, with mean x. The
variance of the sample, denoted by s 2 , is the average of the squared deviations from the sample mean:

s 2 = — -. (13.2)

Since the sample variance is squared, it is also not directly comparable with the mean and the data
themselves.

A common question at this point is "Why is the numerator squared?" One answer is: to get rid of the
negative signs. Numbers are going to fall above and below the mean and, since the variance is looking for
distance, it would be counterproductive if those distances factored each other out.

13.1.2.1.3 Difference between Population Variance and Sample Variance

As seen a distinction is made between the variance, a 2 , of a whole population and the variance, s 2 of a
sample extracted from the population.

When dealing with the complete population the (population) variance is a constant, a parameter which
helps to describe the population. When dealing with a sample from the population the (sample) variance
varies from sample to sample. Its value is only of interest as an estimate for the population variance.

13.1.2.1.4 Properties of Variance

The variance is never negative because the squares are always positive or zero. The unit of variance is the
square of the unit of observation. For example, the variance of a set of heights measured in centimeters will
be given in square centimeters. This fact is inconvenient and has motivated many statisticians to instead
use the square root of the variance, known as the standard deviation, as a summary of dispersion.

13.1.2.2 Standard Deviation

Since the variance is a squared quantity, it cannot be directly compared to the data values or the mean value
of a data set. It is therefore more useful to have a quantity which is the square root of the variance. This
quantity is known as the standard deviation.

163

In statistics, the standard deviation is the most common measure of statistical dispersion. Standard
deviation measures how spread out the values in a data set are. More precisely, it is a measure of the average
distance between the values of the data in the set and the mean. If the data values are all similar, then
the standard deviation will be low (closer to zero). If the data values are highly variable, then the standard
variation is high (further from zero).

The standard deviation is always a positive number and is always measured in the same units as the
original data. For example, if the data are distance measurements in metres, the standard deviation will
also be measured in metres.

13.1.2.2.1 Population Standard Deviation

Let the population consist of n elements {:ei,:e2, ...,x n }, with mean x. The standard deviation of the
population, denoted by a, is the square root of the average of the square of the distance of each data value
from the mean value.

£0>

(13.3)

13.1.2.2.2 Sample Standard Deviation

Let the sample consist of n elements {x\,X2, •■•, x n }, taken from the population, with mean x. The standard
deviation of the sample, denoted by s, is the square root of the average of the squared deviations from the
sample mean:

£(*

i

(13.4)

It is often useful to set your data out in a table so that you can apply the formulae easily. For example to
calculate the standard deviation of {57; 53; 58; 65; 48; 50; 66; 51}, you could set it out in the following way:

sum of items
number of items

n

448

(13.5)

56
Note: To get the deviations, subtract each number from the mean.

X

Deviation (X - X)

Deviation squared [X — X)

57

1

1

53

-3

9

58

2

4

65

9

81

48

-8

64

50

-6

36

66

10

100

51

-5

25

J2 X = 448

2> = o

J2(X-X) 2 = 320

164 CHAPTER 13. STATISTICS

Table 13.1

Note: The sum of the deviations of scores about their mean is zero. This always happens; that is
(X — X) = 0, for any set of data. Why is this? Find out.

Calculate the variance (add the squared results together and divide this total by the number of items).

Variance = — ^ — -

(13.6)

Standard deviation

(13.7)

13.1.2.2.3 Difference between Population Variance and Sample Variance

As with variance, there is a distinction between the standard deviation, a, of a whole population and the
standard deviation, s, of sample extracted from the population.

When dealing with the complete population the (population) standard deviation is a constant, a param-
eter which helps to describe the population. When dealing with a sample from the population the (sample)
standard deviation varies from sample to sample.

In other words, the standard deviation can be calculated as follows:

1. Calculate the mean value x.

2. For each data value Xi calculate the difference Xi — x between xi and the mean value x.

3. Calculate the squares of these differences.

4. Find the average of the squared differences. This quantity is the variance, a 1 .

5. Take the square root of the variance to obtain the standard deviation, a.

Khan academy video on standard deviation

Figure 13.2

Exercise 13.1: Variance and Standard Deviation (Solution on p. 177.)

What is the variance and standard deviation of the population of possibilities associated with
rolling a fair die?

165

13.1.2.3 Interpretation and Application

A large standard deviation indicates that the data values are far from the mean and a small standard
deviation indicates that they are clustered closely around the mean.

For example, each of the three samples (0, 0, 14, 14), (0, 6, 8, 14), and (6, 6, 8, 8) has a mean of 7.
Their standard deviations are 8.08, 5.77 and 1.15, respectively. The third set has a much smaller standard
deviation than the other two because its values are all close to 7. The value of the standard deviation
can be considered 'large' or 'small' only in relation to the sample that is being measured. In this case, a
standard deviation of 7 may be considered large. Given a different sample, a standard deviation of 7 might
be considered small.

Standard deviation may be thought of as a measure of uncertainty. In physical science for example,
the reported standard deviation of a group of repeated measurements should give the precision of those
measurements. When deciding whether measurements agree with a theoretical prediction, the standard
deviation of those measurements is of crucial importance: if the mean of the measurements is too far away
from the prediction (with the distance measured in standard deviations), then we consider the measurements
as contradicting the prediction. This makes sense since they fall outside the range of values that could
reasonably be expected to occur if the prediction were correct and the standard deviation appropriately
quantified. (See prediction interval.)

13.1.2.4 Relationship between Standard Deviation and the Mean

The mean and the standard deviation of a set of data are usually reported together. In a certain sense, the
standard deviation is a "natural" measure of statistical dispersion if the center of the data is measured about
the mean.

13.1.2.4.1 Means and standard deviations

1. Bridget surveyed the price of petrol at petrol stations in Cape Town and Durban. The raw data, in
rands per litre, are given below:

Cape Town

8,96

8,76

9,00

8,91

8,69

8,72

Durban

8,97

8,81

8,52

9,08

8,88

8,68

Table 13.2

a. Find the mean price in each city and then state which city has the lowest mean.

b. Assuming that the data is a population find the standard deviation of each city's prices.

c. Assuming the data is a sample find the standard deviation of each city's prices.

d. Giving reasons which city has the more consistently priced petrol?

2. The following data represents the pocket money of a sample of teenagers. 150; 300; 250; 270; 130; 80;
700; 500; 200; 220; 110; 320; 420; 140. What is the standard deviation?

3. Consider a set of data that gives the weights of 50 cats at a cat show.

a. When is the data seen as a population?

b. When is the data seen as a sample?

4. Consider a set of data that gives the results of 20 pupils in a class.

a. When is the data seen as a population?

b. When is the data seen as a sample?

166 CHAPTER 13. STATISTICS

13.2 Graphical representation of data 2

13.2,1 Graphical Representation of Measures of Central Tendency and Dispersion

The measures of central tendency (mean, median, mode) and the measures of dispersion (range, semi-inter-
quartile range, quartiles, percentiles, inter-quartile range) are numerical methods of summarising data. This
section presents methods of representing the summarised data using graphs.

13.2.1.1 Five Number Summary

One method of summarising a data set is to present a five number summary. The five numbers are: minimum,
first quartile, median, third quartile and maximum.

13.2.1.2 Box and Whisker Diagrams

A box and whisker diagram is a method of depicting the five number summary, graphically.

The main features of the box and whisker diagram are shown in Figure 13.3. The box can lie horizontally
(as shown) or vertically. For a horizonatal diagram, the left edge of the box is placed at the first quartile
and the right edge of the box is placed at the third quartile. The height of the box is arbitrary, as there is
no y-axis. Inside the box there is some representation of central tendency, with the median shown with a
vertical line dividing the box into two. Additionally, a star or asterix is placed at the mean value, centered in
the box in the vertical direction. The whiskers which extend to the sides reach the minimum and maximum
values.

Image notjinished

Figure 13.3

Exercise 13.2: Box and Whisker Diagram (Solution on p. 177.)

Draw a box and whisker diagram for the data set

x = {1,25; 1,5; 2, 5; 2, 5; 3,1; 3, 2; 4,1; 4, 25; 4, 75; 4, 8; 4, 95; 5,1}.

Khan academy video on box and whisker plots

Figure 13.4

13.2.1.2.1 Box and whisker plots

1. Lisa works as a telesales person. She keeps a record of the number of sales she makes each month. The
data below show how much she sells each month. 49; 12; 22; 35; 2; 45; 60; 48; 19; 1; 43; 12 Give
a five number summary and a box and whisker plot of her sales.

2 This content is available online at <http://siyavula.cnx.Org/content/m38860/l.l/>.

167

2. Jason is working in a computer store. He sells the following number of computers each month: 27;
39; 3; 15; 43; 27; 19; 54; 65; 23; 45; 16 Give a five number summary and a box and whisker plot
of his sales,

3. The number of rugby matches attended by 36 season ticket holders is as follows: 15; 11; 7; 34; 24;
22; 31; 12; 912; 9; 1; 3; 15; 5; 8; 11; 225; 2; 6; 18; 16; 17; 20; 13; 1714; 13; 11; 5; 3; 2; 23;
26; 40

a. Sum the data.

b. Using an appropriate graphical method (give reasons) represent the data.

c. Find the median, mode and mean.

d. Calculate the five number summary and make a box and whisker plot.

e. What is the variance and standard deviation?

f. Comment on the data's spread.

g. Where are 95% of the results expected to lie?

4. Rose has worked in a florists shop for nine months. She sold the following number of wedding bouquets:
16; 14; 8; 12; 6; 5; 3; 5; 7

a. What is the five-number summary of the data?

b. Since there is an odd number of data points what do you observe when calculating the five-
numbers?

13.2.1.3 Cumulative Histograms

Cumulative histograms, also known as ogives, are a plot of cumulative frequency and are used to determine
how many data values lie above or below a particular value in a data set. The cumulative frequency is
calculated from a frequency table, by adding each frequency to the total of the frequencies of all data
values before it in the data set. The last value for the cumulative frequency will always be equal to the
total number of data values, since all frequencies will already have been added to the previous total. The
cumulative frequency is plotted at the upper limit of the interval.

For example, the cumulative frequencies for Data Set 2 are shown in Table 13.3 and is drawn in .

Intervals

< n < 1

1 < n < 2

2 < n < 3

3 < n < 4

4 < n < 5

5 < n < 6

Frequency

30

32

35

34

37

32

Cumulative
Frequency

30

30 + 32

30 + 32 + 35

30 + 32 + 35
+ 34

30 + 32 + 35

+ 34 + 37

30 + 32 + 35

+ 34 + 37 +
32

30

62

97

131

168

200

Table 13.3: Cumulative Frequencies for Data Set 2.

Image not finished

Figure 13.5

Notice the frequencies plotted at the upper limit of the intervals, so the points (30;1) (62;2) (97;3), etc
have been plotted. This is different from the frequency polygon where we plot frequencies at the midpoints
of the intervals.

168 CHAPTER 13. STATISTICS

13.2.1.3.1 Intervals

1. Use the following data of peoples ages to answer the questions. 2; 5; 1; 76; 34; 23; 65; 22; 63; 45; 53;
38 4; 28; 5; 73; 80; 17; 15; 5; 34; 37; 45; 56

a. Using an interval width of 8 construct a cumulative frequency distribution

b. How many are below 30?

c. How many are below 60?

d. Giving an explanation state below what value the bottom 50% of the ages fall

e. Below what value do the bottom 40% fall?

f. Construct a frequency polygon and an ogive.

g. Compare these two plots

2. The weights of bags of sand in grams is given below (rounded to the nearest tenth): 50,1; 40,4; 48,5;
29,4; 50,2; 55,3; 58,1; 35,3; 54,2; 43,5 60,1; 43,9; 45,3; 49,2; 36,6; 31,5; 63,1; 49,3; 43,4; 54,1

a. Decide on an interval width and state what you observe about your choice.

d. Construct a cumultative frequency graph and a frequency polygon.

e. Compare the cumulative frequency graph and frequency polygon.

f. Below what value do 53% of the cases fall?

g. Below what value fo 60% of the cases fall?

13.3 Distribution of data 3
13.3.1 Distribution of Data

13.3.1.1 Symmetric and Skewed Data

The shape of a data set is important to know.

Definition 13.1: Shape of a data set

This describes how the data is distributed relative to the mean and median.

• Symmetrical data sets are balanced on either side of the median.

Figure 13.6

Skewed data is spread out on one side more than on the other. It can be skewed right or skewed left.

Image not finished

Figure 13.7

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169

13.3.1.2 Relationship of the Mean, Median, and Mode

The relationship of the mean, median, and mode to each other can provide some information about the
relative shape of the data distribution. If the mean, median, and mode are approximately equal to each
other, the distribution can be assumed to be approximately symmetrical. With both the mean and median
known, the following can be concluded:

• (mean - median) w then the data is symmetrical

• (mean - median) > then the data is positively skewed (skewed to the right). This means that the
median is close to the start of the data set.

• (mean - median) < then the data is negatively skewed (skewed to the left). This means that the
median is close to the end of the data set.

13.3.1.2.1 Distribution of Data

1. Three sets of 12 pupils each had test score recorded. The test was out of 50. Use the given data to

Set 1

Set 2

Set 3

25

32

43

47

34

47

15

35

16

17

32

43

16

25

38

26

16

44

24

38

42

27

47

50

22

43

50

24

29

44

12

18

43

31

25

42

Table 13.4: Cumulative Frequencies for Data Set 2.

a. For each of the sets calculate the mean and the five number summary.

b. For each of the classes find the difference between the mean and the median. Make box and
whisker plots on the same set of axes.

c. State which of the three are skewed (either right or left).

d. Is set A skewed or symmetrical?

e. Is set C symmetrical? Why or why not?

2. Two data sets have the same range and interquartile range, but one is skewed right and the other is
skewed left. Sketch the box and whisker plots and then invent data (6 points in each set) that meets
the requirements.

170

CHAPTER 13. STATISTICS

13.3.2 Scatter Plots

A scatter-plot is a graph that shows the relationship between two variables. We say this is bivariate data
and we plot the data from two different sets using ordered pairs. For example, we could have mass on the
horizontal axis (first variable) and height on the second axis (second variable), or we could have current on
the horizontal axis and voltage on the vertical axis.

Ohm's Law is an important relationship in physics. Ohm's law describes the relationship between current
and voltage in a conductor, like a piece of wire. When we measure the voltage (dependent variable) that
results from a certain current (independent variable) in a wire, we get the data points as shown in Table
13.5.

Current

Voltage

Current

Voltage

0,4

2,4

1,4

0,2

0,3

2,6

1,6

0,4

0,6

2,8

1,9

0,6

0,6

3

1,9

0,8

0,4

3,2

2

1

1

3,4

1,9

1,2

0,9

3,6

2,1

1,4

0,7

3,8

2,1

1,6

1

4

2,4

1,8

1,1

4,2

2,4

2

1,3

4,4

2,5

2,2

1,1

4,6

2,5

Table 13.5: Values of current and voltage measured in a wire.
When we plot this data as points, we get the scatter plot shown in Figure 13.8.

Image not finished

Figure 13.8

If we are to come up with a function that best describes the data, we would have to say that a straight
line best describes this data.

13.3.2.1 Ohm's Law

Ohm's Law describes the relationship between current and voltage in a conductor. The gradient of the graph
of voltage vs. current is known as the resistance of the conductor.

13.3.2.2 Research Project : Scatter Plot

The function that best describes a set of data can take any form. We will restrict ourselves to the forms
already studied, that is, linear, quadratic or exponential. Plot the following sets of data as scatter plots and

171

deduce the type of function that best describes the data. The type of function can either be quadratic or
exponential.

1.

X

y

X

y

X

y

X

y

-5

9,8

14,2

-2,5

11,9

2,5

49,3

-4,5

4,4

0,5

22,5

-2

6,9

3

68,9

-4

7,6

1

21,5

-1,5

8,2

3,5

88,4

-3,5

7,9

1,5

27,5

-1

7,8

4

117,2

-3

7,5

2

41,9

-0,5

14,4

4,5

151,4

2.

3.

Table 13.6

X

y

X

y

X

y

X

y

-5

75

5

-2,5

27,5

2,5

7,5

-4,5

63,5

0,5

3,5

-2

21

3

11

-4

53

1

3

-1,5

15,5

3,5

15,5

-3,5

43,5

1,5

3,5

-1

11

4

21

-3

35

2

5

-0,5

7,5

4,5

27,5

Table 13.7

Height (cm)

147

150

152

155

157

160

163

165

168

170

173

175

178

180

183

Weight (kg)

52

53

54

56

57

59

60

61

63

64

66

68

70

72

74

Table 13.8

Definition 13.2: outlier

A point on a scatter plot which is widely separated from the other points or a result differing
greatly from others in the same sample is called an outlier.

The following simulation allows you to plot scatter plots and fit a curve to the plot. Ignore the error bars
(blue lines) on the points.

Phet simulation for scatter plots

<curve-fitting.swf>

Figure 13.9

172

CHAPTER 13. STATISTICS

13.3.2.3 Scatter Plots

1. A class's results for a test were recorded along with the amount of time spent studying for it. The
results are given below.

Score (percent)

Time spent studying (minutes)

67

100

55

85

70

150

90

180

45

70

75

160

50

80

60

90

84

110

30

60

66

96

96

200

Table 13.9

a. Draw a diagram labelling horizontal and vertical axes.

b. State with reasons, the cause or independent variable and the effect or dependent variable.

c. Plot the data pairs

d. What do you observe about the plot?

e. Is there any pattern emerging?

2. The rankings of eight tennis players is given along with the time they spend practising.

Practice time (min)

Ranking

154

5

390

1

130

6

70

8

240

3

280

2

175

4

103

7

Table 13.10

a. Construct a scatter plot and explain how you chose the dependent (cause) and independent (effect)
variables.

173

b. What pattern or trend do you observe?

3. Eight childrens sweet consumption and sleep habits were recorded. The data is given in the following
table.

Number of sweets (per week)

Average sleeping time (per day)

15

4

12

4,5

5

8

3

8,5

18

3

23

2

11

5

4

8

Table 13.11

a. What is the dependent (cause) variable?

b. What is the independent (effect) variable?

c. Construct a scatter plot of the data.

d. What trend do you observe?

13.4 Misuse of statistics 4
13.4,1 Misuse of Statistics

Statistics can be manipulated in many ways that can be misleading. Graphs need to be carefully analysed
and questions must always be asked about 'the story behind the figures.' Potential manipulations are:

1. Changing the scale to change the appearence of a graph

2. Omissions and biased selection of data

3. Focus on particular research questions

4. Selection of groups

13.4.1.1 Investigation : Misuse of statistics

1. Examine the following graphs and comment on the effects of changing scale.

Image not finished

Figure 13.10

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174

CHAPTER 13. STATISTICS

Image notjtnished

Figure 13.11

2. Examine the following three plots and comment on omission, selection and bias. Hint: What is wrong
with the data and what is missing from the bar and pie charts?

Activity

Hours

Sleep

8

Sports

2

School

7

Visit friend

1

Watch TV

2

Studying

3

Table 13.12

Image notjtnished

Figure 13.12

Image notjtnished

Figure 13.13

13.4.1.2 Misuse of Statistics

The bar graph below shows the results of a study that looked at the cost of food compared to the income of
a household in 1994.

Image notjtnished

Figure 13.14

175

Income (thousands of rands)

Food bill (thousands of rands)

<5

2

5-10

2

10-15

4

15-20

4

20-30

8

30-40

6

40-50

10

> 50

12

Table 13.13

1. What is the dependent variable? Why?

3. What would happen if the graph was changed from food bill in thousands of rands to percentage of
income?

4. Construct this bar graph using a table. What conclusions can be drawn?

5. Why do the two graphs differ despite showing the same information?

6. What else is observed? Does this affect the fairness of the results?

13.4.2 End of Chapter Exercises

1. Many accidents occur during the holidays between Durban and Johannesburg. A study was done to
see if speeding was a factor in the high accident rate. Use the results given to answer the following
questions.

Speed (km/h)

Frequency

60 < x < 70

3

70 < x < 80

2

80 < x < 90

6

90 < x < 100

40

100 < x < 110

50

110 < x < 120

30

120 < x < 130

15

130 < x < 140

12

140 < x < 150

3

150 < x < 160

2

Table 13.14

a. Draw a graph to illustrate this information.

b. Use your graph to find the median speed and the interquartile range.

176 CHAPTER 13. STATISTICS

c. What percent of cars travel more than 120 km.hr 1 on this road?

d. Do cars generally exceed the speed limit?

2. The following two diagrams (showing two schools contribution to charity) have been exaggerated.
Explain how they are misleading and redraw them so that they are not misleading.

Image not finished

Figure 13.15

3. The monthly income of eight teachers are given as follows: RIO 050; R14 300; R9 800; R15 000; R12
140; R13 800; Rll 990; R12 900.

a. What is the mean income and the standard deviation?

b. How many of the salaries are within one standard deviation of the mean?

c. If each teacher gets a bonus of R500 added to their pay what is the new mean and standard
deviation?

d. If each teacher gets a bonus of 10% on their salary what is the new mean and standard deviation?

e. Determine for both of the above, how many salaries are within one standard deviation of the
mean.

f. Using the above information work out which bonus is more beneficial financially for the teachers.

177

Solutions to Exercises in Chapter 13

Solution to Exercise 13.1 (p. 164)

Step 1. When rolling a fair die, the population consists of 6 possible outcomes. The data set is therefore

£ = {1,2,3,4,5,6}. and n=6.
Step 2. The population mean is calculated by:

Step 3. The population variance is calculated by:

±(1 + 2 + 3 + 4 + 5 + 6)
3,5

(13.£

\ (6, 25 + 2, 25 + 0, 25 + 0, 25 + 2, 25 + 6, 25)
2,917

(13.9)

Step 4.

X

(x-x)

(x-x) 2

1

-2.5

6.25

2

-1.5

2.25

3

-0.5

0.25

4

0.5

0.25

5

1.5

2.25

6

2.5

6.25

J2X = 21

J2x =

£ (X-X) 2 = 17.5

Table 13.15

Step 5. The (population) standard deviation is calculated by:

a = ^27917
= 1,708.
Notice how this standard deviation is somewhere in between the possible deviations.
Solution to Exercise 13.2 (p. 166)

Step 1. Minimum =1,25
Maximum = 5,10

Position of first quartile = between 3 and 4
Position of second quartile = between 6 and 7
Position of third quartile = between 9 and 10
Data value between 3 and 4 = I (2, 5 + 2, 5) = 2, 5
Data value between 6 and 7 = \ (3, 2 + 4, 1) = 3, 65
Data value between 9 and 10 = \ (4, 75 + 4, 8) = 4, 775
The five number summary is therefore: 1,25; 2,5; 3,65; 4,775; 5,10.

(13.10)

178 CHAPTER 13. STATISTICS

Image notjtnished

Step 2.

Figure 13.16

Chapter 14

Independent and Dependent Events 1

14.1 Introduction

In probability theory event are either independent or dependent. This chapter describes the differences and
how each type of event is worked with.

14.2 Definitions

Two events are independent if knowing something about the value of one event does not give any information
about the value of the second event. For example, the event of getting a "1" when a die is rolled and the
event of getting a "1" the second time it is thrown are independent.

Definition 14.1: Independent Events

Two events A and B are independent if when one of them happens, it doesn't affect whether the
one happens or not.

The probability of two independent events occurring, P (A D B) , is given by:

P{Ar\B) = P{A) x P{B) (14.1)

Exercise 14.1: Independent Events (Solution on p. 183.)

What is the probability of rolling a 1 and then rolling a 6 on a fair die?

Consequently, two events are dependent if the outcome of the first event affects the outcome of the second
event.

Exercise 14.2: Dependent Events (Solution on p. 183.)

A cloth bag has four coins, one Rl coin, two R2 coins and one R5 coin. What is the probability
of first selecting a Rl coin followed by selecting a R2 coin?

14.2.1 Identification of Independent and Dependent Events

14.2.1.1 Use of a Contingency Table

A two-way contingency table (studied in an earlier grade) can be used to determine whether events are
independent or dependent.

1 This content is available online at <http://siyavula.cnx.Org/content/m32654/l.3/>.

179

180 CHAPTER 14. INDEPENDENT AND DEPENDENT EVENTS

Definition 14.2: two-way contingency table

A two-way contingency table is used to represent possible outcomes when two events are combined
in a statistical analysis.

For example we can draw and analyse a two-way contingency table to solve the following problem.

Exercise 14.3: Contingency Tables (Solution on p. 183.)

A medical trial into the effectiveness of a new medication was carried out. 120 males and 90 females
responded. Out of these 50 males and 40 females responded positively to the medication.

1. Was the medication's success independent of gender? Explain.

2. Give a table for the independence of gender results.

14.2.1.2 Use of a Venn Diagram

We can also use Venn diagrams to check whether events are dependent or independent.

Definition 14.3: Independent events

Events are said to be independent if the result or outcome of one event does not affect the result or
outcome of the other event. So P(A/C)=P(A), where P(A/C) represents the probability of event
A after event C has occured.

Definition 14.4: Dependent events

Two events are dependent if the outcome of one event is affected by the outcome of the other event
i.e. P(A/C)^P(A)

. Also note that P (A/C) = p(c) ■ For example, we can draw a Venn diagram and a contingency table
to illustrate and analyse the following example.

Exercise 14.4: Venn diagrams and events (Solution on p. 184.)

A school decided that its uniform needed upgrading. The colours on offer were beige or blue or
beige and blue. 40% of the school wanted beige, 55% wanted blue and 15% said a combination
would be fine. Are the two events independent?

14.2.1.2.1 Applications of Probability Theory

Two major applications of probability theory in everyday life are in risk assessment and in trade on commodity
markets. Governments typically apply probability methods in environmental regulation where it is called
"pathway analysis", and are often measuring well-being using methods that are stochastic in nature, and
choosing projects to undertake based on statistical analyses of their probable effect on the population as a
whole. It is not correct to say that statistics are involved in the modelling itself, as typically the assessments
of risk are one-time and thus require more fundamental probability models, e.g. "the probability of another
9/11". A law of small numbers tends to apply to all such choices and perception of the effect of such choices,
which makes probability measures a political matter.

A good example is the effect of the perceived probability of any widespread Middle East conflict on oil
prices - which have ripple effects in the economy as a whole. An assessment by a commodity trade that a war
is more likely vs. less likely sends prices up or down, and signals other traders of that opinion. Accordingly,
the probabilities are not assessed independently nor necessarily very rationally. The theory of behavioral
finance emerged to describe the effect of such groupthink on pricing, on policy, and on peace and conflict.

It can reasonably be said that the discovery of rigorous methods to assess and combine probability
assessments has had a profound effect on modern society. A good example is the application of game theory,
itself based strictly on probability, to the Cold War and the mutual assured destruction doctrine. Accordingly,

181

it may be of some importance to most citizens to understand how odds and probability assessments are made,
and how they contribute to reputations and to decisions, especially in a democracy.

Another significant application of probability theory in everyday life is reliability. Many consumer prod-
ucts, such as automobiles and consumer electronics, utilize reliability theory in the design of the product
in order to reduce the probability of failure. The probability of failure is also closely associated with the
product's warranty.

14.3 End of Chapter Exercises

1. In each of the following contingency tables give the expected numbers for the events to be perfectly
independent and decide if the events are independent or dependent.

Brown eyes

Not Brown eyes

Totals

Black hair

50

30

80

Red hair

70

80

150

Totals

120

110

230

Table 14.1

Point A

Point B

Totals

Busses left late

15

40

55

Busses left on time

25

20

45

Totals

40

60

100

Table 14.2

Durban

Bloemfontein

Totals

Liked living there

130

30

160

Did not like living there

140

200

340

Totals

270

230

500

Table 14.3

Multivitamin A

Multivitamin B

Totals

Improvement in health

400

300

700

No improvement in health

140

120

260

Totals

540

420

960

Table 14.4

2. A study was undertaken to see how many people in Port Elizabeth owned either a Volkswagen or a
Toyota. 3% owned both, 25% owned a Toyota and 60% owned a Volkswagen. Draw a contingency
table to show all events and decide if car ownership is independent.

3. Jane invested in the stock market. The probability that she will not lose all her money is 0,32. What
is the probability that she will lose all her money? Explain.

4. If D and F are mutually exclusive events, with P(D')=0,3 and P(D or F)=0,94, find P(F).

5. A car sales person has pink, lime- green and purple models of car A and purple, orange and multicolour
models of car B. One dark night a thief steals a car.

182 CHAPTER 14. INDEPENDENT AND DEPENDENT EVENTS

a. What is the experiment and sample space?

b. Draw a Venn diagram to show this.

c. What is the probability of stealing either a model of A or a model of B?

d. What is the probability of stealing both a model of A and a model of B?

6. The probability of Event X is 0,43 and the probability of Event Y is 0,24. The probability of both
occuring together is 0,10. What is the probability that X or Y will occur (this inculdes X and Y
occuring simultaneously)?

7. P(H)=0,62, P(J)=0,39 and P(H and J)=0,31. Calculate:

a. P(H')

b. P(H or J)

c. P(H' or J')

d. P(H' or J)

e. P(H' and j')

8. The last ten letters of the alphabet were placed in a hat and people were asked to pick one of them.
Event D is picking a vowel, Event E is picking a consonant and Event F is picking the last four letters.
Calculate the following probabilities:

a. P(F')

b. P(F or D)

c. P (neither E nor F)

d. P(D and E)

e. P(E and F)

f. P(E and D')

9. At Dawnview High there are 400 Grade 12's. 270 do Computer Science, 300 do English and 50 do
Typing. All those doing Computer Science do English, 20 take Computer Science and Typing and
35 take English and Typing. Using a Venn diagram calculate the probability that a pupil drawn at
random will take:

a. English, but not Typing or Computer Science

b. English but not Typing

c. English and Typing but not Computer Science

d. English or Typing

183

Solutions to Exercises in Chapter 14

Solution to Exercise 14.1 (p. 179)

Step 1. Event A is rolling a 1 and event B is rolling a 6. Since the outcome of the first event does not affect

the outcome of the second event, the events are independent.
Step 2. The probability of rolling a 1 is | and the probability of rolling a 6 is g.

Therefore, P (A) = \ and P (B) = \.
Step 3.

P{AC\B)

P(A) x P(B)

6 X 6
J_

36

(14.2)

The probability of rolling a 1 and then rolling a 6 on a fair die is gg.
Solution to Exercise 14.2 (p. 179)

Step 1. Event A is selecting a Rl coin and event B is next selecting a R2. Since the outcome of the first event

affects the outcome of the second event (because there are less coins to choose from after the first coin

has been selected), the events are dependent.
Step 2. The probability of first selecting a Rl coin is \ and the probability of next selecting a R2 coin is |

(because after the Rl coin has been selected, there are only three coins to choose from).

Therefore, P (A) = \ and P (B) = §.
Step 3. The same equation as for independent events are used, but the probabilities are calculated differently.

P(AnB)

P{A)x P (B)

I x a

4 X 3

_2_
12

1

6

(14.3)

The probability of first selecting a Rl coin followed by selecting a R2 coin is |.
Solution to Exercise 14.3 (p. 180)
Step 1.

Male

Female

Totals

Positive result

50

40

90

No Positive result

70

50

120

Totals

120

90

210

Table 14.5

Step 2. P (male). P (positive result) = ±f°
P (female). P (positive result) = 21()

21 V

P(male and positive result) =

.mi

0,57
= 0,43
= 0,24

Step 3. P(male and positive result) is the observed probability and P (male) .P (positive result) is the expected
probability. These two are quite different. So there is no evidence that the medication's success is
independent of gender.

Step 4. To get gender independence we need the positive results in the same ratio as the gender. The gender
ratio is: 120:90, or 4:3, so the number in the male and positive column would have to be | of the total
number of patients responding positively which gives 51,4. This leads to the following table:

184

CHAPTER 14. INDEPENDENT AND DEPENDENT EVENTS

Male

Female

Totals

Positive result

51,4

38,6

90

No Positive result

68,6

51,4

120

Totals

120

90

210

Table 14.6

Solution to Exercise 14.4 (p. 180)

Step 1.

Image not finished

Figure 14.1

Step 2.

Beige

Not Beige

Totals

Blue

0,15

0,4

0,55

Not Blue

0,25

0,2

0,35

Totals

0,40

0,6

1

Table 14.7

Step 3. P(Blue)=0,4, P(Beige)=0,55, P(Both)=0,15, P(Neither)=0,20
Probability of choosing beige after blue is:

P (Beige/ Blue)

P(BeigenBlue)

P(Blue)

0,15

0,55

0,27

Step 4. Since P (Beige/ Blue) ^ P (Beige) the events are statistically dependent.

(14.4)

GLOSSARY 185

Glossary

D Dependent events

Two events are dependent if the outcome of one event is affected by the outcome of the other
event i.e. P (A/C) ^ P {A)

I Independent events

Events are said to be independent if the result or outcome of one event does not affect the result
or outcome of the other event. So P(A/C)=P(A), where P(A/C) represents the probability of
event A after event C has occured.

Independent Events

Two events A and B are independent if when one of them happens, it doesn't affect whether the
one happens or not.

M Mathematical Model

A mathematical model is a method of using the mathematical language to describe the
behaviour of a physical system. Mathematical models are used particularly in the natural
sciences and engineering disciplines (such as physics, biology, and electrical engineering) but also
in the social sciences (such as economics, sociology and political science); physicists, engineers,
computer scientists, and economists use mathematical models most extensively.

O outlier

A point on a scatter plot which is widely separated from the other points or a result differing
greatly from others in the same sample is called an outlier.

A quadratic sequence is a sequence of numbers in which the second differences between each
consecutive term differ by the same amount, called a common second difference.

S Shape of a data set

This describes how the data is distributed relative to the mean and median.

T The Area Rule

The area rule applies to any triangle and states that the area of a triangle is given by half the
product of any two sides with the sine of the angle between them.

The Cosine Rule

The cosine rule applies to any triangle and states that:

a 2 = b 2 + c 2 — 2bccos A

b 2 = c 2 + a 2 - Icacos B (12.43)

c 2 = a 2 + b 2 - 2abcos C

186 GLOSSARY

where a is the side opposite A, b is the side opposite B and c is the side opposite C-
The Sine Rule

The sine rule applies to any triangle: ^^ = ^B- = ^^ where a is the side opposite A, b is the

side opposite B and c is the side opposite C-

two-way contingency table

A two-way contingency table is used to represent possible outcomes when two events are
combined in a statistical analysis.

INDEX

187

Index of Keywords and Terms

Keywords are listed by the section with that keyword (page numbers are in parentheses). Keywords
do not necessarily appear in the text of the page. They are merely associated with that section. Ex.
apples, § 1.1 (1) Terms are referenced by the page they appear on. Ex. apples, 1

A algebraic solution, § 4.2(54)

B box-and-whisker plot, § 13.2(166)

C co-ordinate geometry, § 11.3(112)
complete the square, § 2.2(34)
compound depreciation, § 1.2(19)
cone, § 11.1(103)
cosine identity, § 12.5(148)

D Dependent events, 180
depreciation, § 1.4(22)
distribution of data, § 13.3(168)

E effective interest rate, § 1.5(24)
error margins, § (11)
Exponential Functions, § 8(81)
exponents, § (1)

F factorisation, § 2.1(33)

finance, § 1.1(17), § 1.2(19), § 1.4(22),

§ 1-5(24)

Finding i, § 1.4(22)

finding n, § 1.4(22)

finding the equation, § 2.4(39)

five-number summary, § 13.2(166)

future value, § 1.3(21)

G geometry, § 11.1(103), § 11.2(106), § 11.3(112),
§ 11-4(116)

Grade 11, § (1), § (5), § (11), § (13), § 1.1(17),
§ 1.2(19), § 1.3(21), § 1.4(22), § 1.5(24),
§ 2.1(33), § 2.2(34), § 2.3(36), § 2.4(39),
§ 3(47), § 4.1(53), § 4.2(54), § 5(59), § 6(67),
§ 7(75), § 8(81), § 9(87), § 10(93), § 11.1(103),
§ 11.2(106), § 11.3(112), § 11.4(116),
§ 12.1(125), § 12.2(133), § 12.3(137),
§ 12.4(142), § 12.5(148), § 13.1(161),
§ 13.2(166), § 13.3(168), § 13.4(173), § 14(179)
Gradient at a point, § 9(87)
graphical representation, § 13.2(166)
graphical solution, § 4.1(53)
Graphs, § 6(67), § 7(75), § 8(81)

graphs of functions, § 12.1(125)

H histogram, § 13.2(166)

Hyperbolic Functions, § 7(75)

I identities, § 12.2(133)

Independent and dependent events, § 14(179)
Independent Events, 179, 180
interest rate, § 1.5(24)

L Linear Programming, § 10(93)

M Mathematical Model, 59

mathematics, § (1), § (5), § (11), § (13),
§ 5(59), § 9(87), § 12.5(148), § 14(179)
maths, § 1.1(17), § 1.2(19), § 1.3(21),
§ 1.4(22), § 1.5(24), § 2.1(33), § 2.2(34),
§ 2.3(36), § 2.4(39), § 4.1(53), § 4.2(54),
§ 11.1(103), § 11.2(106), § 11.3(112),
§ 11.4(116), § 12.1(125), § 12.2(133),
§ 12.3(137), § 12.4(142), § 13.1(161),
§ 13.2(166), § 13.3(168), § 13.4(173)
misuse of statistics, § 13.4(173)
models, § 5(59)

N nominal interest rate, § 1.5(24)

O outlier, 171

P polygons, § 11.1(103)
population, § 13.1(161)
present value, § 1.3(21)
proportion, § 11.2(106)
pyramid, § 11.1(103)

Q quadratic equation, § 2.1(33), § 2.4(39)
quadratic equations, § 2.2(34), § 2.3(36)

R reduction formulae, § 12.3(137)

188

INDEX

S sample, § 13.1(161)

Shape of a data set, 168

similarity, § 11.1(103)

simple depreciation, § 1.1(17)

simultaneous equation, § 4.1(53), § 4.2(54)

sine identity, § 12.5(148)

skewed data, § 13.3(168)

South Africa, § (1), § (5), § (11), § (13),

§ 1.1(17), § 1.2(19), § 1.3(21), § 1.4(22),

§ 1.5(24), § 2.1(33), § 2.2(34), § 2.3(36),

§ 2.4(39), § 3(47), § 4.1(53), § 4.2(54), § 5(59),

§ 6(67), § 7(75), § 8(81), § 9(87), § 10(93),

§ 11.1(103), § 11.2(106), § 11.3(112),

§ 11.4(116), § 12.1(125), § 12.2(133),

§ 12.3(137), § 12.4(142), § 12.5(148),

§ 13.1(161), § 13.2(166), § 13.3(168),

§ 13.4(173), § 14(179)

sphere, § 11.1(103)

standard deviation, § 13.1(161)

statistics, § 13.1(161), § 13.2(166), § 13.3(168),

§ 13.4(173)

Surds, § (5)

surface area, § 11.1(103)

T The Area Rule, 151
The Cosine Rule, 149
The Sine Rule, 148
transformations, § 11.4(116)
triangle geometry, § 11.2(106)
trigonometric equations, § 12.4(142)
trigonometry, § 12.1(125), § 12.2(133),
§ 12.3(137), § 12.4(142), § 12.5(148)
two-way contingency table, 180

V

variance,
volume, i

§ 13.1(161)
11.1(103)

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URL: http://siyavula.cnx.org/content/m38873/l-l/

Pages: 148-154

Copyright: Free High School Science Texts Project

Module: "Statistics: Standard deviation and variance"

Used here as: "Standard deviation and variance"

By: Free High School Science Texts Project

URL: http://siyavula.cnx.org/content/m38858/l-l/

Pages: 161-165

Copyright: Free High School Science Texts Project

Module: "Statistics: graphical representation of data"

Used here as: "Graphical representation of data"

By: Free High School Science Texts Project

URL: http://siyavula.cnx.org/content/m38860/l-l/

Pages: 166-168

Copyright: Free High School Science Texts Project

Module: "Statistics: distribution of data"

Used here as: "Distribution of data"

By: Free High School Science Texts Project

URL: http://siyavula.cnx.org/content/m38861/l-l/

Pages: 168-173

Copyright: Free High School Science Texts Project

Module: "Statistics: misuse of statistics"

Used here as: "Misuse of statistics"

By: Free High School Science Texts Project

URL: http://siyavula.cnx.org/content/m38864/l-l/

Pages: 173-176

Copyright: Free High School Science Texts Project

Module: "Independent and Dependent Events"

By: Rory Adams, Free High School Science Texts Project, Sarah Blyth, Heather Williams

URL: http://siyavula.cnx.org/content/m32654/l-3/

Pages: 179-184

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